编辑距离 | DP-5

给定两个字符串str1 和 str2 以及可以在 str1 上执行的操作。查找将“str1”转换为“str2”所需的最少编辑(操作)次数。

1. 插入
2. 消除
3. 代替

以上所有操作都是等价的。

例子：

Input:   str1 = "geek", str2 = "gesek"
Output:  1
We can convert str1 into str2 by inserting a 's'.

Input:   str1 = "cat", str2 = "cut"
Output:  1
We can convert str1 into str2 by replacing 'a' with 'u'.

Input:   str1 = "sunday", str2 = "saturday"
Output:  3
Last three and first characters are same.  We basically
need to convert "un" to "atur".  This can be done using
below three operations. 
Replace 'n' with 'r', insert t, insert a

这种情况下的子问题是什么?
这个想法是从两个字符串的左侧或右侧开始一个一个地处理所有字符。
让我们从右角开始遍历，每对被遍历的字符都有两种可能性。

m: Length of str1 (first string)
n: Length of str2 (second string)

1. 如果两个字符串的最后一个字符相同，则没什么可做的。忽略最后一个字符并获取剩余字符串的计数。所以我们重复长度为 m-1 和 n-1。
2. 否则(如果最后一个字符不同)，我们考虑对 ‘str1’ 的所有操作，考虑对第一个字符串 的最后一个字符的所有三个操作，递归计算所有三个操作的最小成本并取三个值的最小值。
   1. 插入：重复 m 和 n-1
   2. 删除：重复 m-1 和 n
   3. 替换：对 m-1 和 n-1 重复

以下是上述朴素递归解决方案的实现。

C++

// A Naive recursive C++ program to find minimum number
// operations to convert str1 to str2
#include 
using namespace std;
 
// Utility function to find minimum of three numbers
int min(int x, int y, int z) { return min(min(x, y), z); }
 
int editDist(string str1, string str2, int m, int n)
{
    // If first string is empty, the only option is to
    // insert all characters of second string into first
    if (m == 0)
        return n;
 
    // If second string is empty, the only option is to
    // remove all characters of first string
    if (n == 0)
        return m;
 
    // If last characters of two strings are same, nothing
    // much to do. Ignore last characters and get count for
    // remaining strings.
    if (str1[m - 1] == str2[n - 1])
        return editDist(str1, str2, m - 1, n - 1);
 
    // If last characters are not same, consider all three
    // operations on last character of first string,
    // recursively compute minimum cost for all three
    // operations and take minimum of three values.
    return 1
           + min(editDist(str1, str2, m, n - 1), // Insert
                 editDist(str1, str2, m - 1, n), // Remove
                 editDist(str1, str2, m - 1,
                          n - 1) // Replace
             );
}
 
// Driver code
int main()
{
    // your code goes here
    string str1 = "sunday";
    string str2 = "saturday";
 
    cout << editDist(str1, str2, str1.length(),
                     str2.length());
 
    return 0;
}

Java

// A Naive recursive Java program to find minimum number
// operations to convert str1 to str2
class EDIST {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
 
    static int editDist(String str1, String str2, int m,
                        int n)
    {
        // If first string is empty, the only option is to
        // insert all characters of second string into first
        if (m == 0)
            return n;
 
        // If second string is empty, the only option is to
        // remove all characters of first string
        if (n == 0)
            return m;
 
        // If last characters of two strings are same,
        // nothing much to do. Ignore last characters and
        // get count for remaining strings.
        if (str1.charAt(m - 1) == str2.charAt(n - 1))
            return editDist(str1, str2, m - 1, n - 1);
 
        // If last characters are not same, consider all
        // three operations on last character of first
        // string, recursively compute minimum cost for all
        // three operations and take minimum of three
        // values.
        return 1
            + min(editDist(str1, str2, m, n - 1), // Insert
                  editDist(str1, str2, m - 1, n), // Remove
                  editDist(str1, str2, m - 1,
                           n - 1) // Replace
              );
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String str1 = "sunday";
        String str2 = "saturday";
 
        System.out.println(editDist(
            str1, str2, str1.length(), str2.length()));
    }
}
/*This code is contributed by Rajat Mishra*/

Python

# A Naive recursive Python program to fin minimum number
# operations to convert str1 to str2
 
 
def editDistance(str1, str2, m, n):
 
    # If first string is empty, the only option is to
    # insert all characters of second string into first
    if m == 0:
        return n
 
    # If second string is empty, the only option is to
    # remove all characters of first string
    if n == 0:
        return m
 
    # If last characters of two strings are same, nothing
    # much to do. Ignore last characters and get count for
    # remaining strings.
    if str1[m-1] == str2[n-1]:
        return editDistance(str1, str2, m-1, n-1)
 
    # If last characters are not same, consider all three
    # operations on last character of first string, recursively
    # compute minimum cost for all three operations and take
    # minimum of three values.
    return 1 + min(editDistance(str1, str2, m, n-1),    # Insert
                   editDistance(str1, str2, m-1, n),    # Remove
                   editDistance(str1, str2, m-1, n-1)    # Replace
                   )
 
 
# Driver code
str1 = "sunday"
str2 = "saturday"
print editDistance(str1, str2, len(str1), len(str2))
 
# This code is contributed by Bhavya Jain

C#

// A Naive recursive C# program to
// find minimum numberoperations
// to convert str1 to str2
using System;
 
class GFG {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
 
    static int editDist(String str1, String str2, int m,
                        int n)
    {
        // If first string is empty, the only option is to
        // insert all characters of second string into first
        if (m == 0)
            return n;
 
        // If second string is empty, the only option is to
        // remove all characters of first string
        if (n == 0)
            return m;
 
        // If last characters of two strings are same,
        // nothing much to do. Ignore last characters and
        // get count for remaining strings.
        if (str1[m - 1] == str2[n - 1])
            return editDist(str1, str2, m - 1, n - 1);
 
        // If last characters are not same, consider all
        // three operations on last character of first
        // string, recursively compute minimum cost for all
        // three operations and take minimum of three
        // values.
        return 1
            + min(editDist(str1, str2, m, n - 1), // Insert
                  editDist(str1, str2, m - 1, n), // Remove
                  editDist(str1, str2, m - 1,
                           n - 1) // Replace
              );
    }
 
    // Driver code
    public static void Main()
    {
        String str1 = "sunday";
        String str2 = "saturday";
        Console.WriteLine(
            editDist(str1, str2, str1.Length, str2.Length));
    }
}
 
// This Code is Contributed by Sam007

Javascript

 
// Javascript program to
// find minimum numberoperations
// to convert str1 to str2
function min(x, y, z)
{
    if (x <= y && x <= z)
        return x;
    if (y <= x && y <= z)
        return y;
    else
        return z;
}
 
function editDist(str1, str2, m, n)
{
     
    // If first string is empty, the
    // only option is to insert all
    // characters of second string leto first
    if (m == 0)
        return n;
 
    // If second string is empty, the only
    // option is to remove all characters
    // of first string
    if (n == 0)
        return m;
 
    // If last characters of two strings are
    // same, nothing much to do. Ignore last
    // characters and get count for remaining
    // strings.
    if (str1[m - 1] == str2[n - 1])
        return editDist(str1, str2, m - 1, n - 1);
 
    // If last characters are not same, consider all
    // three operations on last character of first
    // string, recursively compute minimum cost for all
    // three operations and take minimum of three
    // values.
    return 1 +
    min(editDist(str1, str2, m, n - 1), // Insert
        editDist(str1, str2, m - 1, n), // Remove
        editDist(str1, str2, m - 1, n - 1)); // Replace
}
 
// Driver code
let str1 = "sunday";
let str2 = "saturday";
document.write(editDist(str1, str2, str1.length,
                                    str2.length));
 
// This code is contributed by target_2
 

---

输出

3

上述解决方案的时间复杂度是指数级的。在最坏的情况下，我们最终可能会进行 O(3 ^m ) 次操作。最坏的情况发生在两个字符串的字符都不匹配时。下面是最坏情况的递归调用图。

我们可以看到很多子问题都解决了，一次又一次，例如eD(2, 2)被调用了3次。由于再次调用相同的子问题，因此该问题具有重叠子问题的属性。所以编辑距离问题具有动态规划问题的两个属性(见this和this)。与其他典型的动态规划 (DP) 问题一样，可以通过构造一个存储子问题结果的临时数组来避免对相同子问题的重新计算。

C++

// A Dynamic Programming based C++ program to find minimum
// number operations to convert str1 to str2
#include 
using namespace std;
 
// Utility function to find the minimum of three numbers
int min(int x, int y, int z) { return min(min(x, y), z); }
 
int editDistDP(string str1, string str2, int m, int n)
{
    // Create a table to store results of subproblems
    int dp[m + 1][n + 1];
 
    // Fill d[][] in bottom up manner
    for (int i = 0; i <= m; i++) {
        for (int j = 0; j <= n; j++) {
            // If first string is empty, only option is to
            // insert all characters of second string
            if (i == 0)
                dp[i][j] = j; // Min. operations = j
 
            // If second string is empty, only option is to
            // remove all characters of second string
            else if (j == 0)
                dp[i][j] = i; // Min. operations = i
 
            // If last characters are same, ignore last char
            // and recur for remaining string
            else if (str1[i - 1] == str2[j - 1])
                dp[i][j] = dp[i - 1][j - 1];
 
            // If the last character is different, consider
            // all possibilities and find the minimum
            else
                dp[i][j]
                    = 1
                      + min(dp[i][j - 1], // Insert
                            dp[i - 1][j], // Remove
                            dp[i - 1][j - 1]); // Replace
        }
    }
 
    return dp[m][n];
}
 
// Driver code
int main()
{
    // your code goes here
    string str1 = "sunday";
    string str2 = "saturday";
 
    cout << editDistDP(str1, str2, str1.length(),
                       str2.length());
 
    return 0;
}

Java

// A Dynamic Programming based Java program to find minimum
// number operations to convert str1 to str2
class EDIST {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
 
    static int editDistDP(String str1, String str2, int m,
                          int n)
    {
        // Create a table to store results of subproblems
        int dp[][] = new int[m + 1][n + 1];
 
        // Fill d[][] in bottom up manner
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                // If first string is empty, only option is
                // to insert all characters of second string
                if (i == 0)
                    dp[i][j] = j; // Min. operations = j
 
                // If second string is empty, only option is
                // to remove all characters of second string
                else if (j == 0)
                    dp[i][j] = i; // Min. operations = i
 
                // If last characters are same, ignore last
                // char and recur for remaining string
                else if (str1.charAt(i - 1)
                         == str2.charAt(j - 1))
                    dp[i][j] = dp[i - 1][j - 1];
 
                // If the last character is different,
                // consider all possibilities and find the
                // minimum
                else
                    dp[i][j] = 1
                               + min(dp[i][j - 1], // Insert
                                     dp[i - 1][j], // Remove
                                     dp[i - 1]
                                       [j - 1]); // Replace
            }
        }
 
        return dp[m][n];
    }
 
    // Driver Code
    public static void main(String args[])
    {
        String str1 = "sunday";
        String str2 = "saturday";
        System.out.println(editDistDP(
            str1, str2, str1.length(), str2.length()));
    }
} /*This code is contributed by Rajat Mishra*/

Python

# A Dynamic Programming based Python program for edit
# distance problem
 
 
def editDistDP(str1, str2, m, n):
    # Create a table to store results of subproblems
    dp = [[0 for x in range(n + 1)] for x in range(m + 1)]
 
    # Fill d[][] in bottom up manner
    for i in range(m + 1):
        for j in range(n + 1):
 
            # If first string is empty, only option is to
            # insert all characters of second string
            if i == 0:
                dp[i][j] = j    # Min. operations = j
 
            # If second string is empty, only option is to
            # remove all characters of second string
            elif j == 0:
                dp[i][j] = i    # Min. operations = i
 
            # If last characters are same, ignore last char
            # and recur for remaining string
            elif str1[i-1] == str2[j-1]:
                dp[i][j] = dp[i-1][j-1]
 
            # If last character are different, consider all
            # possibilities and find minimum
            else:
                dp[i][j] = 1 + min(dp[i][j-1],        # Insert
                                   dp[i-1][j],        # Remove
                                   dp[i-1][j-1])    # Replace
 
    return dp[m][n]
 
 
# Driver code
str1 = "sunday"
str2 = "saturday"
 
print(editDistDP(str1, str2, len(str1), len(str2)))
# This code is contributed by Bhavya Jain

C＃

// A Dynamic Programming based
// C# program to find minimum
// number operations to
// convert str1 to str2
using System;
 
class GFG {
    static int min(int x, int y, int z)
    {
        if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
    }
 
    static int editDistDP(String str1, String str2, int m,
                          int n)
    {
        // Create a table to store
        // results of subproblems
        int[, ] dp = new int[m + 1, n + 1];
 
        // Fill d[][] in bottom up manner
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                // If first string is empty, only option is
                // to insert all characters of second string
                if (i == 0)
 
                    // Min. operations = j
                    dp[i, j] = j;
 
                // If second string is empty, only option is
                // to remove all characters of second string
                else if (j == 0)
 
                    // Min. operations = i
                    dp[i, j] = i;
 
                // If last characters are same, ignore last
                // char and recur for remaining string
                else if (str1[i - 1] == str2[j - 1])
                    dp[i, j] = dp[i - 1, j - 1];
 
                // If the last character is different,
                // consider all possibilities and find the
                // minimum
                else
                    dp[i, j] = 1
                               + min(dp[i, j - 1], // Insert
                                     dp[i - 1, j], // Remove
                                     dp[i - 1,
                                        j - 1]); // Replace
            }
        }
 
        return dp[m, n];
    }
 
    // Driver code
    public static void Main()
    {
        String str1 = "sunday";
        String str2 = "saturday";
        Console.Write(editDistDP(str1, str2, str1.Length,
                                 str2.Length));
    }
}
// This Code is Contributed by Sam007

Javascript

 
// A Dynamic Programming based
// Javascript program to find minimum
// number operations to convert str1 to str2
 
function min(x,y,z)
{
    if (x <= y && x <= z)
            return x;
        if (y <= x && y <= z)
            return y;
        else
            return z;
}
 
function editDistDP(str1,str2,m,n)
{
    // Create a table to store results of subproblems
        let dp = new Array(m + 1);
        for(let i=0;i<m+1;i++)
        {
            dp[i]=new Array(n+1);
            for(let j=0;j<n+1;j++)
            {
                dp[i][j]=0;
            }
        }
  
        // Fill d[][] in bottom up manner
        for (let i = 0; i <= m; i++) {
            for (let j = 0; j <= n; j++) {
                // If first string is empty, only option is
                // to insert all characters of second string
                if (i == 0)
                    dp[i][j] = j; // Min. operations = j
  
                // If second string is empty, only option is
                // to remove all characters of second string
                else if (j == 0)
                    dp[i][j] = i; // Min. operations = i
  
                // If last characters are same, ignore last
                // char and recur for remaining string
                else if (str1[i - 1]
                         == str2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1];
  
                // If the last character is different,
                // consider all possibilities and find the
                // minimum
                else
                    dp[i][j] = 1
                               + min(dp[i][j - 1], // Insert
                                     dp[i - 1][j], // Remove
                                     dp[i - 1]
                                       [j - 1]); // Replace
            }
        }
  
        return dp[m][n];
}
 
// Driver Code
let str1 = "sunday";
let str2 = "saturday";
document.write(editDistDP(
str1, str2, str1.length, str2.length));
 
// This code is contributed by unknown2108
 

输出

3

时间复杂度： O(mxn)
辅助空间： O(mxn)

空间复杂解决方案：在上面给出的方法中，我们需要 O(mxn) 空间。如果字符串的长度大于 2000，这将不合适，因为它只能创建 2000 x 2000 的二维数组。要填充 DP 数组中的一行，我们只需要上一行。例如，如果我们在 DP 数组中填充 i = 10 行，我们只需要第 9 行的值。所以我们简单地创建一个 2 x str1 长度的 DP 数组。这种方法降低了空间复杂度。下面是上述问题的C++实现

C++

// A Space efficient Dynamic Programming
// based C++ program to find minimum
// number operations to convert str1 to str2
#include 
using namespace std;
 
void EditDistDP(string str1, string str2)
{
    int len1 = str1.length();
    int len2 = str2.length();
 
    // Create a DP array to memoize result
    // of previous computations
    int DP[2][len1 + 1];
 
    // To fill the DP array with 0
    memset(DP, 0, sizeof DP);
 
    // Base condition when second string
    // is empty then we remove all characters
    for (int i = 0; i <= len1; i++)
        DP[0][i] = i;
 
    // Start filling the DP
    // This loop run for every
    // character in second string
    for (int i = 1; i <= len2; i++) {
        // This loop compares the char from
        // second string with first string
        // characters
        for (int j = 0; j <= len1; j++) {
            // if first string is empty then
            // we have to perform add character
            // operation to get second string
            if (j == 0)
                DP[i % 2][j] = i;
 
            // if character from both string
            // is same then we do not perform any
            // operation . here i % 2 is for bound
            // the row number.
            else if (str1[j - 1] == str2[i - 1]) {
                DP[i % 2][j] = DP[(i - 1) % 2][j - 1];
            }
 
            // if character from both string is
            // not same then we take the minimum
            // from three specified operation
            else {
                DP[i % 2][j] = 1 + min(DP[(i - 1) % 2][j],
                                       min(DP[i % 2][j - 1],
                                           DP[(i - 1) % 2][j - 1]));
            }
        }
    }
 
    // after complete fill the DP array
    // if the len2 is even then we end
    // up in the 0th row else we end up
    // in the 1th row so we take len2 % 2
    // to get row
    cout << DP[len2 % 2][len1] << endl;
}
 
// Driver program
int main()
{
    string str1 = "food";
    string str2 = "money";
    EditDistDP(str1, str2);
    return 0;
}

Java

// A Space efficient Dynamic Programming
// based Java program to find minimum
// number operations to convert str1 to str2
import java.util.*;
class GFG
{
 
static void EditDistDP(String str1, String str2)
{
    int len1 = str1.length();
    int len2 = str2.length();
 
    // Create a DP array to memoize result
    // of previous computations
    int [][]DP = new int[2][len1 + 1];
 
 
    // Base condition when second String
    // is empty then we remove all characters
    for (int i = 0; i <= len1; i++)
        DP[0][i] = i;
 
    // Start filling the DP
    // This loop run for every
    // character in second String
    for (int i = 1; i <= len2; i++)
    {
       
        // This loop compares the char from
        // second String with first String
        // characters
        for (int j = 0; j <= len1; j++)
        {
           
            // if first String is empty then
            // we have to perform add character
            // operation to get second String
            if (j == 0)
                DP[i % 2][j] = i;
 
            // if character from both String
            // is same then we do not perform any
            // operation . here i % 2 is for bound
            // the row number.
            else if (str1.charAt(j - 1) == str2.charAt(i - 1)) {
                DP[i % 2][j] = DP[(i - 1) % 2][j - 1];
            }
 
            // if character from both String is
            // not same then we take the minimum
            // from three specified operation
            else {
                DP[i % 2][j] = 1 + Math.min(DP[(i - 1) % 2][j],
                                       Math.min(DP[i % 2][j - 1],
                                           DP[(i - 1) % 2][j - 1]));
            }
        }
    }
 
    // after complete fill the DP array
    // if the len2 is even then we end
    // up in the 0th row else we end up
    // in the 1th row so we take len2 % 2
    // to get row
    System.out.print(DP[len2 % 2][len1] +"\n");
}
 
// Driver program
public static void main(String[] args)
{
    String str1 = "food";
    String str2 = "money";
    EditDistDP(str1, str2);
}
}
 
// This code is contributed by aashish1995

蟒蛇3

# A Space efficient Dynamic Programming
# based Python3 program to find minimum
# number operations to convert str1 to str2
def EditDistDP(str1, str2):
     
    len1 = len(str1)
    len2 = len(str2)
 
    # Create a DP array to memoize result
    # of previous computations
    DP = [[0 for i in range(len1 + 1)]
             for j in range(2)];
 
    # Base condition when second String
    # is empty then we remove all characters
    for i in range(0, len1 + 1):
        DP[0][i] = i
 
    # Start filling the DP
    # This loop run for every
    # character in second String
    for i in range(1, len2 + 1):
         
        # This loop compares the char from
        # second String with first String
        # characters
        for j in range(0, len1 + 1):
 
            # If first String is empty then
            # we have to perform add character
            # operation to get second String
            if (j == 0):
                DP[i % 2][j] = i
 
            # If character from both String
            # is same then we do not perform any
            # operation . here i % 2 is for bound
            # the row number.
            elif(str1[j - 1] == str2[i-1]):
                DP[i % 2][j] = DP[(i - 1) % 2][j - 1]
             
            # If character from both String is
            # not same then we take the minimum
            # from three specified operation
            else:
                DP[i % 2][j] = (1 + min(DP[(i - 1) % 2][j],
                                    min(DP[i % 2][j - 1],
                                  DP[(i - 1) % 2][j - 1])))
             
    # After complete fill the DP array
    # if the len2 is even then we end
    # up in the 0th row else we end up
    # in the 1th row so we take len2 % 2
    # to get row
    print(DP[len2 % 2][len1], "")
 
# Driver code
if __name__ == '__main__':
     
    str1 = "food"
    str2 = "money"
     
    EditDistDP(str1, str2)
 
# This code is contributed by gauravrajput1

C＃

// A Space efficient Dynamic Programming
// based C# program to find minimum
// number operations to convert str1 to str2
using System;
class GFG
{
 
static void EditDistDP(String str1, String str2)
{
    int len1 = str1.Length;
    int len2 = str2.Length;
 
    // Create a DP array to memoize result
    // of previous computations
    int [,]DP = new int[2, len1 + 1];
 
 
    // Base condition when second String
    // is empty then we remove all characters
    for (int i = 0; i <= len1; i++)
        DP[0, i] = i;
 
    // Start filling the DP
    // This loop run for every
    // character in second String
    for (int i = 1; i <= len2; i++)
    {
       
        // This loop compares the char from
        // second String with first String
        // characters
        for (int j = 0; j <= len1; j++)
        {
           
            // if first String is empty then
            // we have to perform add character
            // operation to get second String
            if (j == 0)
                DP[i % 2, j] = i;
 
            // if character from both String
            // is same then we do not perform any
            // operation . here i % 2 is for bound
            // the row number.
            else if (str1[j - 1] == str2[i - 1])
            {
                DP[i % 2, j] = DP[(i - 1) % 2, j - 1];
            }
 
            // if character from both String is
            // not same then we take the minimum
            // from three specified operation
            else {
                DP[i % 2, j] = 1 + Math.Min(DP[(i - 1) % 2, j],
                                       Math.Min(DP[i % 2, j - 1],
                                           DP[(i - 1) % 2, j - 1]));
            }
        }
    }
 
    // after complete fill the DP array
    // if the len2 is even then we end
    // up in the 0th row else we end up
    // in the 1th row so we take len2 % 2
    // to get row
    Console.Write(DP[len2 % 2, len1] +"\n");
}
 
// Driver program
public static void Main(String[] args)
{
    String str1 = "food";
    String str2 = "money";
    EditDistDP(str1, str2);
}
}
 
// This code is contributed by aashish1995

Javascript

// A Space efficient Dynamic Programming
// based Javascript program to find minimum
// number operations to convert str1 to str2
function EditDistDP(str1, str2)
{
    let len1 = str1.length;
    let len2 = str2.length;
  
    // Create a DP array to memoize result
    // of previous computations
    let DP = new Array(2);
    for(let i = 0; i < 2; i++)
    {
        DP[i] = new Array(len1+1);
        for(let j = 0; j < len1 + 1; j++)
            DP[i][j] = 0;
    }
  
  
    // Base condition when second String
    // is empty then we remove all characters
    for (let i = 0; i <= len1; i++)
        DP[0][i] = i;
  
    // Start filling the DP
    // This loop run for every
    // character in second String
    for (let i = 1; i <= len2; i++)
    {
        
        // This loop compares the char from
        // second String with first String
        // characters
        for (let j = 0; j <= len1; j++)
        {
            
            // if first String is empty then
            // we have to perform add character
            // operation to get second String
            if (j == 0)
                DP[i % 2][j] = i;
  
            // if character from both String
            // is same then we do not perform any
            // operation . here i % 2 is for bound
            // the row number.
            else if (str1[j-1] == str2[i-1]) {
                DP[i % 2][j] = DP[(i - 1) % 2][j - 1];
            }
  
            // if character from both String is
            // not same then we take the minimum
            // from three specified operation
            else {
                DP[i % 2][j] = 1 + Math.min(DP[(i - 1) % 2][j],
                                       Math.min(DP[i % 2][j - 1],
                                           DP[(i - 1) % 2][j - 1]));
            }
        }
    }
  
    // after complete fill the DP array
    // if the len2 is even then we end
    // up in the 0th row else we end up
    // in the 1th row so we take len2 % 2
    // to get row
    document.write(DP[len2 % 2][len1] +"<br>");
}
 
// Driver program
let str1 = "food";
let str2 = "money";
EditDistDP(str1, str2);
 
// This code is contributed by patel2127.

输出

4

时间复杂度： O(mxn)
辅助空间： O(m)

这是递归的记忆版本，即自顶向下 DP：

C++14

#include 
using namespace std;
int minDis(string s1, string s2, int n, int m, vector> &dp){
         
  // If any string is empty,
  // return the remaining characters of other string
           
  if(n==0)    return m;
         
  if(m==0)    return n;
         
           
  // To check if the recursive tree
  // for given n & m has already been executed
         
  if(dp[n][m]!=-1)    return dp[n][m];
           
           
  // If characters are equal, execute
  // recursive function for n-1, m-1
         
  if(s1[n-1]==s2[m-1]){          
    if(dp[n-1][m-1]==-1){              
      return dp[n][m] = minDis(s1, s2, n-1, m-1, dp);          
    }       
    else
      return dp[n][m] = dp[n-1][m-1];  
  }
   
         
  // If characters are nt equal, we need to
           
  // find the minimum cost out of all 3 operations.
         
  else{          
    int m1, m2, m3;        // temp variables  
     
    if(dp[n-1][m]!=-1){   
      m1 = dp[n-1][m];     
    }          
    else{  
      m1 = minDis(s1, s2, n-1, m, dp);     
    }           
             
    if(dp[n][m-1]!=-1){               
      m2 = dp[n][m-1];           
    }           
    else{   
      m2 = minDis(s1, s2, n, m-1, dp);     
    }                                  
    
    if(dp[n-1][m-1]!=-1){   
      m3 = dp[n-1][m-1];     
    }  
    else{  
      m3 = minDis(s1, s2, n-1, m-1, dp);      
    }    
    return dp[n][m] = 1+min(m1, min(m2, m3));       
  }
   
}
 
 
// Driver program
int main() {
     
  string str1 = "voldemort";
  string str2 = "dumbledore";
       
  int n= str1.length(), m = str2.length();   
  vector> dp(n+1, vector(m+1, -1));
                 
  cout<

Java

import java.util.*;
class GFG
{
   
static int minDis(String s1, String s2,
                  int n, int m, int[][]dp)
{
         
  // If any String is empty,
  // return the remaining characters of other String
  if(n == 0)   
    return m; 
  if(m == 0)   
    return n;
              
  // To check if the recursive tree
  // for given n & m has already been executed
  if(dp[n][m] != -1)   
    return dp[n][m];
                 
  // If characters are equal, execute
  // recursive function for n-1, m-1
  if(s1.charAt(n - 1) == s2.charAt(m - 1))
  {          
    if(dp[n - 1][m - 1] == -1)
    {              
      return dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp);          
    }       
    else
      return dp[n][m] = dp[n - 1][m - 1];  
  }
   
  // If characters are nt equal, we need to
           
  // find the minimum cost out of all 3 operations.     
  else
  {          
    int m1, m2, m3;        // temp variables  
    if(dp[n-1][m] != -1)
    {   
      m1 = dp[n - 1][m];     
    }          
    else
    {  
      m1 = minDis(s1, s2, n - 1, m, dp);     
    }           
             
    if(dp[n][m - 1] != -1)
    {               
      m2 = dp[n][m - 1];           
    }           
    else
    {   
      m2 = minDis(s1, s2, n, m - 1, dp);     
    }                                  
    
    if(dp[n - 1][m - 1] != -1)
    {   
      m3 = dp[n - 1][m - 1];     
    }  
    else
    {  
      m3 = minDis(s1, s2, n - 1, m - 1, dp);      
    }    
    return dp[n][m] = 1 + Math.min(m1, Math.min(m2, m3));       
  }
}
 
// Driver program
public static void main(String[] args)
{
     
  String str1 = "voldemort";
  String str2 = "dumbledore";
       
  int n= str1.length(), m = str2.length();   
  int[][] dp = new int[n + 1][m + 1];
  for(int i = 0; i < n + 1; i++)
  Arrays.fill(dp[i], -1);               
  System.out.print(minDis(str1, str2, n, m, dp));
}
}
 
// This code is contributed by gauravrajput1

蟒蛇3

def minDis(s1, s2, n, m, dp) :
          
  # If any string is empty,
  # return the remaining characters of other string         
  if(n == 0) :
      return m       
  if(m == 0) :
      return n
                    
  # To check if the recursive tree
  # for given n & m has already been executed
  if(dp[n][m] != -1)  :
      return dp[n][m];
                   
  # If characters are equal, execute
  # recursive function for n-1, m-1   
  if(s1[n - 1] == s2[m - 1]) :          
    if(dp[n - 1][m - 1] == -1) :
        dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp)
        return dp[n][m]                  
    else :
        dp[n][m] = dp[n - 1][m - 1]
        return dp[n][m]
         
  # If characters are nt equal, we need to          
  # find the minimum cost out of all 3 operations.        
  else :           
    if(dp[n - 1][m] != -1) :  
      m1 = dp[n - 1][m]     
    else :
      m1 = minDis(s1, s2, n - 1, m, dp)
              
    if(dp[n][m - 1] != -1) :               
      m2 = dp[n][m - 1]           
    else :
      m2 = minDis(s1, s2, n, m - 1, dp)  
    if(dp[n - 1][m - 1] != -1) :   
      m3 = dp[n - 1][m - 1]   
    else :
      m3 = minDis(s1, s2, n - 1, m - 1, dp)
     
    dp[n][m] = 1 + min(m1, min(m2, m3))
    return dp[n][m]
     
    # Driver code
str1 = "voldemort"
str2 = "dumbledore"
    
n = len(str1)
m = len(str2)
dp = [[-1 for i in range(m + 1)] for j in range(n + 1)]
              
print(minDis(str1, str2, n, m, dp))
 
# This code is contributed by divyesh072019.

C＃

using System;
using System.Collections.Generic;
class GFG {
 
  static int minDis(string s1, string s2, int n,
                    int m, List> dp)
  {
 
    // If any string is empty,
    // return the remaining characters of other string
    if(n == 0)  
      return m;
 
    if(m == 0)   
      return n;
 
    // To check if the recursive tree
    // for given n & m has already been executed
    if(dp[n][m] != -1) 
      return dp[n][m];
 
    // If characters are equal, execute
    // recursive function for n-1, m-1
    if(s1[n - 1] == s2[m - 1])
    {          
      if(dp[n - 1][m - 1] == -1)
      {              
        return dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp);          
      }       
      else
        return dp[n][m] = dp[n - 1][m - 1];  
    }
 
 
    // If characters are nt equal, we need to
    // find the minimum cost out of all 3 operations.
    else
    {          
      int m1, m2, m3;        // temp variables  
 
      if(dp[n - 1][m] != -1)
      {   
        m1 = dp[n - 1][m];     
      }          
      else
      {  
        m1 = minDis(s1, s2, n - 1, m, dp);     
      }           
 
      if(dp[n][m - 1] != -1)
      {               
        m2 = dp[n][m - 1];           
      }           
      else
      {   
        m2 = minDis(s1, s2, n, m - 1, dp);     
      }                                  
 
      if(dp[n - 1][m - 1] != -1)
      {   
        m3 = dp[n - 1][m - 1];     
      }  
      else
      {  
        m3 = minDis(s1, s2, n - 1, m - 1, dp);      
      }    
      return dp[n][m] = 1+ Math.Min(m1, Math.Min(m2, m3));       
    }
 
  }
 
  // Driver code
  static void Main()
  {
    string str1 = "voldemort";
    string str2 = "dumbledore";
 
    int n = str1.Length, m = str2.Length;  
    List> dp = new List>();
    for(int i = 0; i < n + 1; i++)
    {
      dp.Add(new List());
      for(int j = 0; j < m + 1; j++)
      {
        dp[i].Add(-1);
      }
    }
    Console.WriteLine(minDis(str1, str2, n, m, dp));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

 
function minDis(s1,s2,n,m,dp)
{
    // If any String is empty,
  // return the remaining characters of other String
  if(n == 0)  
    return m;
  if(m == 0)  
    return n;
               
  // To check if the recursive tree
  // for given n & m has already been executed
  if(dp[n][m] != -1)  
    return dp[n][m];
                  
  // If characters are equal, execute
  // recursive function for n-1, m-1
  if(s1[n - 1] == s2[m - 1])
  {         
    if(dp[n - 1][m - 1] == -1)
    {             
      return dp[n][m] = minDis(s1, s2, n - 1, m - 1, dp);         
    }      
    else
      return dp[n][m] = dp[n - 1][m - 1]; 
  }
    
  // If characters are nt equal, we need to
            
  // find the minimum cost out of all 3 operations.    
  else
  {         
    let m1, m2, m3;        // temp variables 
    if(dp[n-1][m] != -1)
    {  
      m1 = dp[n - 1][m];    
    }         
    else
    { 
      m1 = minDis(s1, s2, n - 1, m, dp);    
    }          
              
    if(dp[n][m - 1] != -1)
    {              
      m2 = dp[n][m - 1];          
    }          
    else
    {  
      m2 = minDis(s1, s2, n, m - 1, dp);    
    }                                 
     
    if(dp[n - 1][m - 1] != -1)
    {  
      m3 = dp[n - 1][m - 1];    
    } 
    else
    { 
      m3 = minDis(s1, s2, n - 1, m - 1, dp);     
    }   
    return dp[n][m] = 1 + Math.min(m1, Math.min(m2, m3));      
  }
}
 
// Driver program
 
let str1 = "voldemort";
let str2 = "dumbledore";
 
let n= str1.length, m = str2.length;  
let dp = new Array(n + 1);
for(let i = 0; i < n + 1; i++)
{
    dp[i]=new Array(m+1);
    for(let j=0;j<m+1;j++)
        dp[i][j]=-1;
}
 
document.write(minDis(str1, str2, n, m, dp));
 
// This code is contributed by avanitrachhadiya2155
 

输出

7

应用：编辑距离算法有很多实际应用，参考 Lucene API 示例。另一个示例，显示字典中与给定单词拼写错误的单词非常接近的所有单词。

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。