LeetCode:Edit Distance

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本文介绍了解决两个字符串间编辑距离问题的动态规划方法。通过插入、删除或替换字符操作，找到将一个字符串转换为另一个字符串所需的最小步骤数，并提供了详细的算法实现。

Edit Distance

Total Accepted: 58999 Total Submissions: 204641 Difficulty: Hard

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

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(M) One Edit Distance

思路：动规

设：word1="abcd"，word2="bbc"。

则其状态转移过程为：

Ø a b c d
Ø 0 1 2 3 4
b 1 1 1 2 3
b 2 2 1 2 3
c 3 3 2 1 2

其状态转移方程为：

dp[i][j] = dp[i-1][j-1],	if word1[i-1]==word2[j-1];
dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1,	otherwise;

c++ code:

/*
  Ø a b c d
Ø 0 1 2 3 4
b 1 1 1 2 3
b 2 2 1 2 3
c 3 3 2 1 2
*/

class Solution {
public:
    int minDistance(string word1, string word2) {
        
        int m = word1.length();
        int n = word2.length();
        
        int dp[m+1][n+1];
        
        for(int i=0;i<=m;i++) dp[i][0] = i; // 初始化第一列
        for(int j=0;j<=n;j++) dp[0][j] = j; // 初始化第一行
        
        for(int i=1;i<=m;i++) {
            for(int j=1;j<=n;j++) {
                
                if(word1[i-1]==word2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = min3(dp[i-1][j-1],dp[i-1][j],dp[i][j-1]) + 1;
                
            }
        }
        return dp[m][n];
    }
    
    // 自定义函数
    int min3(int x, int y,int z) {
        return min(x,min(y,z));
    }
};