打印最长公共子序列

给定两个序列，打印它们中存在的最长子序列。
例子：
输入序列“ABCDGH”和“AEDFHR”的 LCS 是长度为 3 的“ADH”。
输入序列“AGGTAB”和“GXTXAYB”的 LCS 是长度为 4 的“GTAB”。
我们在上一篇文章中讨论了最长公共子序列（LCS）问题。那里讨论的函数主要是找到 LCS 的长度。为了找到 LCS 的长度，构建了一个 2D 表 L[][]。在这篇文章中，讨论了构建和打印 LCS 的函数。
以下是打印 LCS 的详细算法。它使用相同的 2D 表 L[][]。
1)使用上一篇文章中讨论的步骤构造 L[m+1][n+1]。
2)值 L[m][n] 包含 LCS 的长度。创建一个长度等于 lcs 长度加 1 的字符数组 lcs[]（存储 \0 的额外一个）。
2)从 L[m][n] 开始遍历二维数组。对每个单元格 L[i][j] 执行以下操作
..... a)如果对应于 L[i][j] 的字符（在 X 和 Y 中）相同（或 X[i-1] == Y[j-1]），则将此字符作为 LCS 的一部分包含在内.
..... b)否则比较 L[i-1][j] 和 L[i][j-1] 的值，然后朝着更大的值方向前进。
下表（取自 Wiki）显示了上述算法的步骤（突出显示）。

		0	1	2	3	4	5	6	7
		Ø	M	Z	J	A	W	X	U
0	Ø	0	0	0	0	0	0	0	0
1	X	0	0	0	0	0	0	1	1
2	M	0	1	1	1	1	1	1	1
3	J	0	1	1	2	2	2	2	2
4	Y	0	1	1	2	2	2	2	2
5	A	0	1	1	2	3	3	3	3
6	U	0	1	1	2	3	3	3	4
7	Z	0	1	2	2	3	3	3	4

以下是上述方法的实现。

C++

/* Dynamic Programming implementation of LCS problem */
#include
#include
#include
using namespace std;
 
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
void lcs( char *X, char *Y, int m, int n )
{
  int L[m+1][n+1];
 
  /* Following steps build L[m+1][n+1] in bottom up fashion. Note
    that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
  for (int i=0; i<=m; i++)
  {
    for (int j=0; j<=n; j++)
    {
      if (i == 0 || j == 0)
        L[i][j] = 0;
      else if (X[i-1] == Y[j-1])
        L[i][j] = L[i-1][j-1] + 1;
      else
        L[i][j] = max(L[i-1][j], L[i][j-1]);
    }
  }
 
  // Following code is used to print LCS
  int index = L[m][n];
 
  // Create a character array to store the lcs string
  char lcs[index+1];
  lcs[index] = '\0'; // Set the terminating character
 
  // Start from the right-most-bottom-most corner and
  // one by one store characters in lcs[]
  int i = m, j = n;
  while (i > 0 && j > 0)
  {
    // If current character in X[] and Y are same, then
    // current character is part of LCS
    if (X[i-1] == Y[j-1])
    {
      lcs[index-1] = X[i-1]; // Put current character in result
      i--; j--; index--;     // reduce values of i, j and index
    }
 
    // If not same, then find the larger of two and
    // go in the direction of larger value
    else if (L[i-1][j] > L[i][j-1])
      i--;
    else
      j--;
  }
 
  // Print the lcs
  cout << "LCS of " << X << " and " << Y << " is " << lcs;
}
 
/* Driver program to test above function */
int main()
{
  char X[] = "AGGTAB";
  char Y[] = "GXTXAYB";
  int m = strlen(X);
  int n = strlen(Y);
  lcs(X, Y, m, n);
  return 0;
}

Java

// Dynamic Programming implementation of LCS problem in Java
import java.io.*;
 
class  LongestCommonSubsequence
{
    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    static void lcs(String X, String Y, int m, int n)
    {
        int[][] L = new int[m+1][n+1];
  
        // Following steps build L[m+1][n+1] in bottom up fashion. Note
        // that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
        for (int i=0; i<=m; i++)
        {
            for (int j=0; j<=n; j++)
            {
                if (i == 0 || j == 0)
                    L[i][j] = 0;
                else if (X.charAt(i-1) == Y.charAt(j-1))
                    L[i][j] = L[i-1][j-1] + 1;
                else
                    L[i][j] = Math.max(L[i-1][j], L[i][j-1]);
            }
        }
  
        // Following code is used to print LCS
        int index = L[m][n];
        int temp = index;
  
        // Create a character array to store the lcs string
        char[] lcs = new char[index+1];
        lcs[index] = '\u0000'; // Set the terminating character
  
        // Start from the right-most-bottom-most corner and
        // one by one store characters in lcs[]
        int i = m;
        int j = n;
        while (i > 0 && j > 0)
        {
            // If current character in X[] and Y are same, then
            // current character is part of LCS
            if (X.charAt(i-1) == Y.charAt(j-1))
            {
                // Put current character in result
                lcs[index-1] = X.charAt(i-1);
                 
                // reduce values of i, j and index
                i--;
                j--;
                index--;    
            }
  
            // If not same, then find the larger of two and
            // go in the direction of larger value
            else if (L[i-1][j] > L[i][j-1])
                i--;
            else
                j--;
        }
  
        // Print the lcs
        System.out.print("LCS of "+X+" and "+Y+" is ");
        for(int k=0;k<=temp;k++)
            System.out.print(lcs[k]);
    }
     
    // driver program
    public static void main (String[] args)
    {
        String X = "AGGTAB";
        String Y = "GXTXAYB";
        int m = X.length();
        int n = Y.length();
        lcs(X, Y, m, n);
    }
}
 
// Contributed by Pramod Kumar

Python3

# Dynamic programming implementation of LCS problem
 
# Returns length of LCS for X[0..m-1], Y[0..n-1]
def lcs(X, Y, m, n):
    L = [[0 for x in range(n+1)] for x in range(m+1)]
 
    # Following steps build L[m+1][n+1] in bottom up fashion. Note
    # that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1]
    for i in range(m+1):
        for j in range(n+1):
            if i == 0 or j == 0:
                L[i][j] = 0
            else if X[i-1] == Y[j-1]:
                L[i][j] = L[i-1][j-1] + 1
            else:
                L[i][j] = max(L[i-1][j], L[i][j-1])
 
    # Following code is used to print LCS
    index = L[m][n]
 
    # Create a character array to store the lcs string
    lcs = [""] * (index+1)
    lcs[index] = ""
 
    # Start from the right-most-bottom-most corner and
    # one by one store characters in lcs[]
    i = m
    j = n
    while i > 0 and j > 0:
 
        # If current character in X[] and Y are same, then
        # current character is part of LCS
        if X[i-1] == Y[j-1]:
            lcs[index-1] = X[i-1]
            i-=1
            j-=1
            index-=1
 
        # If not same, then find the larger of two and
        # go in the direction of larger value
        else if L[i-1][j] > L[i][j-1]:
            i-=1
        else:
            j-=1
 
    print ("LCS of " + X + " and " + Y + " is " + "".join(lcs) )
 
# Driver program
X = "AGGTAB"
Y = "GXTXAYB"
m = len(X)
n = len(Y)
lcs(X, Y, m, n)
 
# This code is contributed by BHAVYA JAIN

C#

// Dynamic Programming implementation
// of LCS problem in C#
using System;
 
class GFG
{
    // Returns length of LCS for X[0..m-1], Y[0..n-1]
    static void lcs(String X, String Y, int m, int n)
    {
        int[,] L = new int[m+1, n+1];
 
        // Following steps build L[m+1][n+1] in
        // bottom up fashion. Note that L[i][j]
        // contains length of LCS of X[0..i-1]
        // and Y[0..j-1]
        for (int i = 0; i <= m; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                if (i == 0 || j == 0)
                    L[i, j] = 0;
                else if (X[i-1] == Y[j-1])
                    L[i, j] = L[i-1, j-1] + 1;
                else
                    L[i, j] = Math.Max(L[i-1, j], L[i, j-1]);
            }
        }
 
        // Following code is used to print LCS
        int index = L[m, n];
        int temp = index;
 
        // Create a character array
        // to store the lcs string
        char[] lcs = new char[index+1];
         
        // Set the terminating character
        lcs[index] = '\0';
 
        // Start from the right-most-bottom-most corner
        // and one by one store characters in lcs[]
        int k = m, l = n;
        while (k > 0 && l > 0)
        {
            // If current character in X[] and Y
            // are same, then current character
            // is part of LCS
            if (X[k-1] == Y[l-1])
            {
                // Put current character in result
                lcs[index-1] = X[k-1];
                 
                // reduce values of i, j and index
                k--;
                l--;
                index--;    
            }
 
            // If not same, then find the larger of two and
            // go in the direction of larger value
            else if (L[k-1, l] > L[k, l-1])
                k--;
            else
                l--;
        }
 
        // Print the lcs
        Console.Write("LCS of " + X + " and " + Y + " is ");
        for(int q = 0; q <= temp; q++)
            Console.Write(lcs[q]);
    }
     
    // Driver program
    public static void Main ()
    {
        String X = "AGGTAB";
        String Y = "GXTXAYB";
        int m = X.Length;
        int n = Y.Length;
        lcs(X, Y, m, n);
    }
}
 
// This code is contributed by Sam007

PHP

 0 && $j > 0)
    {
        // If current character in X[] and Y are same,
        // then current character is part of LCS
        if ($X[$i - 1] == $Y[$j - 1])
        {
            // Put current character in result
            $lcs[$index - 1] = $X[$i - 1];
            $i--;
            $j--;
            $index--;    // reduce values of i, j and index
        }
     
        // If not same, then find the larger of two
        // and go in the direction of larger value
        else if ($L[$i - 1][$j] > $L[$i][$j - 1])
            $i--;
        else
            $j--;
    }
     
    // Print the lcs
    echo "LCS of " . $X . " and " . $Y . " is ";
    for($k = 0; $k < $temp; $k++)
        echo $lcs[$k];
}
 
// Driver Code
$X = "AGGTAB";
$Y = "GXTXAYB";
$m = strlen($X);
$n = strlen($Y);
lcs($X, $Y, $m, $n);
 
// This code is contributed by ita_c
?>

Javascript

function ReverseString(str) {
   return str.split('').reverse().join('')
}
   
function max(a, b)
{
    if (a > b)
        return a;
    else
        return b;
}
function printLCS(str1, str2) {
    var len1 = str1.length;
    var len2 = str2.length;
    var lcs = new Array(len1 + 1);
    for (var i = 0; i <= len1; i++) {
        lcs[i] = new Array(len2 + 1)
    }
    for (var i = 0; i <= len1; i++) {
        for (var j = 0; j <= len2; j++) {
            if (i == 0 || j == 0) {
                lcs[i][j] = 0;
            }
            else {
                if (str1[i - 1] == str2[j - 1]) {
                    lcs[i][j] = 1 + lcs[i - 1][j - 1];
                }
                else {
                    lcs[i][j] = max(lcs[i][j - 1], lcs[i - 1][j]);
                }
            }
        }}
        
        var n = lcs[len1][len2];
         document.write("Length of common subsequence is: " +
         n + "<br>" + "The subsequence is : ");
        var str="";
       var i = len1;
       var j = len2;
       while(i>0&&j>0)
       {
            if(str1[i - 1] == str2[j - 1])
            {
                str += str1[i - 1];
                i--;
                j--;
            }
            else{
            if(lcs[i][j-1]>lcs[i-1][j])
            {
                j--;
            }
            else
            {
                i--;
            }
            }
        }
       return ReverseString(str);
    }
    var str1 = "AGGTAB";
    var str2 = "GXTXAYB";
    document.write(printLCS(str1, str2));
     
    // This code is contributed by akshitsaxenaa09

输出：

LCS of AGGTAB and GXTXAYB is GTAB