字符串单词之间的最小距离

给定一个字符串s以及 S 中存在的两个单词w1和w2 。任务是找到w1和w2之间的最小距离。这里，距离是第一个词和第二个词之间的步数或词数。

例子：

Input : s = “geeks for geeks contribute practice”, w1 = “geeks”, w2 = “practice” 
Output : 1 
There is only one word between the closest occurrences of w1 and w2.

Input : s = “the quick the brown quick brown the frog”, w1 = “quick”, w2 = “frog”
Output : 2

一个简单的方法是考虑 w1 的每次出现。对于 w1 的每次出现，找到最近的 w2 并跟踪最小距离。

C++

// C++ program to find Minimum Distance
// Between Words of a String
#include 
#include 
using namespace std;
 
// Function to implement split function
void split(const string &s, char delimiter,
                     vector &words)
{
    string token;
    stringstream tokenStream(s);
 
    while (getline(tokenStream, token, delimiter))
        words.push_back(token);
}
 
// Function to calculate the minimum
// distance between w1 and w2 in s
int distance(string s, string w1, string w2)
{
    if (w1 == w2)
        return 0;
 
    // get individual words in a list
    vector words;
    split(s, ' ', words);
 
    // assume total length of the string
    // as minimum distance
    int min_dist = words.size() + 1;
 
    // traverse through the entire string
    for (int index = 0; index < words.size(); index++)
    {
        if (words[index] == w1)
        {
            for (int search = 0;
                     search < words.size(); search++)
            {
                if (words[search] == w2)
                {
                    // the distance between the words is
                    // the index of the first word - the
                    // current word index
                    int curr = abs(index - search) - 1;
 
                    // comparing current distance with
                    // the previously assumed distance
                    if (curr < min_dist)
                        min_dist = curr;
                }
            }
        }
    }
 
    // w1 and w2 are same and adjacent
    return min_dist;
}
 
// Driver Code
int main(int argc, char const *argv[])
{
    string s = "geeks for geeks contribute practice";
    string w1 = "geeks";
    string w2 = "practice";
    cout << distance(s, w1, w2) << endl;
    return 0;
}
 
// This code is contributed by
// sanjeev2552

Java

// Java program to find Minimum Distance
// Between Words of a String
 class solution
 {
 
// Function to calculate the minimum
// distance between w1 and w2 in s
static int distance(String s,String w1,String w2)
{
      
    if (w1 .equals( w2) )
        return 0 ;
  
    // get individual words in a list
    String words[] = s.split(" ");
  
    // assume total length of the string
    // as minimum distance
    int min_dist = (words.length) + 1;
  
    // traverse through the entire string
    for (int index = 0;
         index < words.length ; index ++)
    {
  
        if (words[index] .equals( w1))
        {
            for (int search = 0;
                 search < words.length; search ++)
            {
                if (words[search] .equals(w2))
                {
                    // the distance between the words is
                    // the index of the first word - the
                    // current word index
                    int curr = Math.abs(index - search) - 1;
  
                    // comparing current distance with
                    // the previously assumed distance
                    if (curr < min_dist)
                    {
                        min_dist = curr ;
                    }
                }
            }
        }
    }
      
    // w1 and w2 are same and adjacent
    return min_dist;
}
  
// Driver code
public static void main(String args[])
{
     
String s = "geeks for geeks contribute practice";
String w1 = "geeks" ;
String w2 = "practice" ;
  
System.out.print( distance(s, w1, w2) );
  
}
}
//contributed by Arnab Kundu

Python3

# function to calculate the minimum
# distance between w1 and w2 in s
  
def distance(s, w1, w2):
     
    if w1 == w2 :
       return 0
 
    # get individual words in a list
    words = s.split(" ")
 
    # assume total length of the string as
    # minimum distance
    min_dist = len(words)+1
 
    # traverse through the entire string
    for index in range(len(words)):
 
        if words[index] == w1:
            for search in range(len(words)):
 
                if words[search] == w2:
 
                    # the distance between the words is
                    # the index of the first word - the
                    # current word index
                    curr = abs(index - search) - 1;
 
                    # comparing current distance with
                    # the previously assumed distance
                    if curr < min_dist:
                       min_dist = curr
 
    # w1 and w2 are same and adjacent
    return min_dist
     
 
# Driver code
s = "geeks for geeks contribute practice"
w1 = "geeks"
w2 = "practice"
print(distance(s, w1, w2))

C#

// C# program to find Minimum Distance
// Between Words of a String
 
using System;
 class solution
 {
  
// Function to calculate the minimum
// distance between w1 and w2 in s
static int distance(string s,string w1,string w2)
{
       
    if (w1 .Equals( w2) )
        return 0 ;
   
    // get individual words in a list
    string[] words = s.Split(" ");
   
    // assume total length of the string
    // as minimum distance
    int min_dist = (words.Length) + 1;
   
    // traverse through the entire string
    for (int index = 0;
         index < words.Length ; index ++)
    {
   
        if (words[index] .Equals( w1))
        {
            for (int search = 0;
                 search < words.Length; search ++)
            {
                if (words[search] .Equals(w2))
                {
                    // the distance between the words is
                    // the index of the first word - the
                    // current word index
                    int curr = Math.Abs(index - search) - 1;
   
                    // comparing current distance with
                    // the previously assumed distance
                    if (curr < min_dist)
                    {
                        min_dist = curr ;
                    }
                }
            }
        }
    }
       
    // w1 and w2 are same and adjacent
    return min_dist;
}
   
// Driver code
public static void Main()
{
      
string s = "geeks for geeks contribute practice";
string w1 = "geeks" ;
string w2 = "practice" ;
   
Console.Write( distance(s, w1, w2) );
   
}
}

Javascript

// Javascript program to find Minimum Distance
// Between Words of a String
     
// Function to calculate the minimum
// distance between w1 and w2 in s
function distance(s,w1,w2) {
    if (w1 ==( w2) )
        return 0 ;
   
    // get individual words in a list
    let words = s.split(" ");
   
    // assume total length of the string
    // as minimum distance
    let min_dist = (words.length) + 1;
   
    // traverse through the entire string
    for (let index = 0;
         index < words.length ; index ++)
    {
   
        if (words[index] == ( w1))
        {
            for (let search = 0;
                 search < words.length; search ++)
            {
                if (words[search] == (w2))
                {
                    // the distance between the words is
                    // the index of the first word - the
                    // current word index
                    let curr = Math.abs(index - search) - 1;
   
                    // comparing current distance with
                    // the previously assumed distance
                    if (curr < min_dist)
                    {
                        min_dist = curr ;
                    }
                }
            }
        }
    }
       
    // w1 and w2 are same and adjacent
    return min_dist;
}
     
    // Driver code
    let s = "geeks for geeks contribute practice";
    let w1 = "geeks" ;
    let w2 = "practice" ;
    document.write( distance(s, w1, w2) );
     
     
     
// This code is contributed by rag2127

---

---

输出：

1

一个有效的解决方案是找到任何元素的第一次出现，然后跟踪前一个元素和当前元素。如果它们不同并且距离小于当前最小值，则更新最小值。

C++

// C++ program to extract words from
// a string using stringstream
#include
using namespace std;
 
int distance(string s, string w1, string w2)
{
 
  if (w1 == w2)
  {
    return 0;
  }
 
  vector words;
 
  // Used to split string around spaces.
  istringstream ss(s);
 
  string word; // for storing each word
 
  // Traverse through all words
  // while loop till we get
  // strings to store in string word
  while (ss >> word)
  {
    words.push_back(word);
  }
 
  int n = words.size();
 
  // assume total length of the string as
  // minimum distance
  int min_dist = n + 1;
 
  // Find the first occurrence of any of the two
  // numbers (w1 or w2) and store the index of
  // this occurrence in prev
  int prev = 0, i = 0;
 
  for (i = 0; i < n; i++)
  {
 
    if (words[i] == w1 || (words[i] == w2))
    {
      prev = i;
      break;
    }
  }
 
  // Traverse after the first occurrence
  while (i < n)
  {
    if (words[i] == w1 || (words[i] == w2))
    {
 
      // If the current element matches with
      // any of the two then check if current
      // element and prev element are different
      // Also check if this value is smaller than
      // minimum distance so far
      if ((words[prev] != words[i]) &&
          (i - prev) < min_dist)
      {
        min_dist = i - prev - 1;
        prev = i;
      }
      else
      {
        prev = i;
      }
    }
    i += 1;
 
  }
  return min_dist;
}
 
// Driver code
int main()
{
  string s = "geeks for geeks contribute practice";
  string w1 = "geeks";
  string w2 = "practice";
  cout<

Java

// Java program to extract words from
// a string using stringstream
class GFG {
 
    static int distance(String s, String w1, String w2) {
 
        if (w1.equals(w2)) {
            return 0;
        }
 
       // get individual words in a list
        String[] words = s.split(" ");
        int n = words.length;
 
        // assume total length of the string as
        // minimum distance
        int min_dist = n + 1;
 
        // Find the first occurrence of any of the two
        // numbers (w1 or w2) and store the index of
        // this occurrence in prev
        int prev = 0, i = 0;
        for (i = 0; i < n; i++) {
 
            if (words[i].equals(w1) || words[i].equals(w2)) {
                prev = i;
                break;
            }
        }
        // Traverse after the first occurrence
        while (i < n) {
            if (words[i].equals(w1) || words[i].equals(w2)) {
 
                // If the current element matches with
                // any of the two then check if current
                // element and prev element are different
                // Also check if this value is smaller than
                // minimum distance so far
                if ((!words[prev].equals(words[i])) && (i - prev) < min_dist) {
                    min_dist = i - prev - 1;
                    prev = i;
                } else {
                    prev = i;
                }
 
            }
            i += 1;
 
        }
        return min_dist;
    }
// Driver code
 
    public static void main(String[] args) {
        String s = "geeks for geeks contribute practice";
        String w1 = "geeks";
        String w2 = "practice";
        System.out.println(distance(s, w1, w2));
// This code is contributed by princiRaj1992
    }
}

C＃

// C# program to extract words from
// a string using stringstream
using System;
 
class GFG
{
 
    static int distance(String s, String w1, String w2)
    {
 
        if (w1.Equals(w2))
        {
            return 0;
        }
 
        // get individual words in a list
        String[] words = s.Split(" ");
        int n = words.Length;
 
        // assume total length of the string as
        // minimum distance
        int min_dist = n + 1;
 
        // Find the first occurrence of any of the two
        // numbers (w1 or w2) and store the index of
        // this occurrence in prev
        int prev = 0, i = 0;
        for (i = 0; i < n; i++)
        {
 
            if (words[i].Equals(w1) || words[i].Equals(w2))
            {
                prev = i;
                break;
            }
        }
         
        // Traverse after the first occurrence
        while (i < n)
        {
            if (words[i].Equals(w1) || words[i].Equals(w2))
            {
 
                // If the current element matches with
                // any of the two then check if current
                // element and prev element are different
                // Also check if this value is smaller than
                // minimum distance so far
                if ((!words[prev].Equals(words[i])) &&
                                (i - prev) < min_dist)
                {
                    min_dist = i - prev - 1;
                    prev = i;
                }
                else
                {
                    prev = i;
                }
            }
            i += 1;
 
        }
        return min_dist;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        String s = "geeks for geeks contribute practice";
        String w1 = "geeks";
        String w2 = "practice";
        Console.Write(distance(s, w1, w2));
    }
}
 
// This code is contributed by Mohit kumar 29

Python3

# Python3 program to extract words from
# a string using stringstream
 
def distance(s, w1, w2):
     
    if w1 == w2 :
       return 0
 
    # get individual words in a list
    words = s.split(" ")
    n = len(words)
 
    # assume total length of the string as
    # minimum distance
    min_dist = n+1
 
    # Find the first occurrence of any of the two
    # numbers (w1 or w2) and store the index of
    # this occurrence in prev
    for i in range(n):
         
        if words[i] == w1 or words[i] == w2:
            prev = i
            break
 
    # Traverse after the first occurrence
    while i < n:
        if words[i] == w1 or words[i] == w2:
 
            # If the current element matches with
            # any of the two then check if current
            # element and prev element are different
            # Also check if this value is smaller than
            # minimum distance so far
            if words[prev] != words[i] and (i - prev) < min_dist :
                min_dist = i - prev - 1
                prev = i
            else:
                prev = i
        i += 1       
 
    return min_dist
 
# Driver code
s = "geeks for geeks contribute practice"
w1 = "geeks"
w2 = "practice"
print(distance(s, w1, w2))

Javascript

 
// Javascript program to extract words from
// a string using stringstream
 
function distance(s,w1,w2)
{
    if (w1 == (w2))
    {
            return 0;
        }
  
       // get individual words in a list
        let words = s.split(" ");
        let n = words.length;
  
        // assume total length of the string as
        // minimum distance
        let min_dist = n + 1;
  
        // Find the first occurrence of any of the two
        // numbers (w1 or w2) and store the index of
        // this occurrence in prev
        let prev = 0, i = 0;
        for (i = 0; i < n; i++)
        {
  
            if (words[i] == (w1) || words[i] == (w2))
            {
                prev = i;
                break;
            }
        }
        // Traverse after the first occurrence
        while (i < n)
        {
            if (words[i] == (w1) || words[i] == (w2))
            {
  
                // If the current element matches with
                // any of the two then check if current
                // element and prev element are different
                // Also check if this value is smaller than
                // minimum distance so far
                if ((words[prev] != (words[i])) && (i - prev) < min_dist) {
                    min_dist = i - prev - 1;
                    prev = i;
                } else {
                    prev = i;
                }
  
            }
            i += 1;
        }
        return min_dist;
}
 
// Driver code
let s = "geeks for geeks contribute practice";
let w1 = "geeks";
let w2 = "practice";
document.write(distance(s, w1, w2));
     
// This code is contributed by avanitrachhadiya2155

输出：

1

一个有效的解决方案是，如果 word1 再次出现，则将 word1 的索引存储在 (lastpos) 变量中，然后如果 word1 不出现，我们更新 (lastpos)，然后简单地找到 word1 和 word2 索引的差异。

C++

// C++ program to find Minimum Distance
// Between Words of a String
#include 
using namespace std;
 
int shortestDistance(vector &s, string word1, string word2)
    {
         if(word1==word2) return  0;
        int ans = INT_MAX;
       //To store the lastposition of word1
        int lastPos = -1;
        for(int i = 0 ; i < s.size() ; i++)
        {
            if(s[i] == word1 || s[i] == word2)
            {
               //first occurrence of word1
                if(lastPos == -1)
                   lastPos = i;
                else
                {
                    //if word1 repeated again we store the last position of word1
                    if(s[lastPos]==s[i])
                     lastPos = i;
                    else
                    {
                      //find the difference of position of word1 and word2
                        ans = min(ans , (i-lastPos)-1);
                        lastPos = i;
                    }
                }
            }
        }
        return ans;
}
//Driver code
int main() {
    vector s{"geeks", "for", "geeks", "contribute",
     "practice"};
     string w1 = "geeks";
    string w2 = "practice";
   
    cout<