[Leetcode] Edit Distance

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

一道经典的DP题目，我们用dp[i][j]表示将word1长度0...i-1和word2长度0...j-1变成相同字符串的所需的最少操作次数。那么有以下几种情况:

1. word1[i-1] == word2[j-1]，那么dp[i][j] = dp[i-1][j-1]
2. 否则，dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])+1.其中，dp[i-1][j-1]代表替换操作，dp[i-1][j]代表插入操作，dp[i][j-1]代表删除操作

    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();
        int[][] dp = new int[m+1][n+1];
        
        for(int i = 0; i <= n; i++)
            dp[0][i] = i;
        for(int i = 0; i <= m; i++)
            dp[i][0] = i;
        for(int i = 1; i <= m; i++) {
            for(int j = 1; j <=n ;j++) {
                if(word1.charAt(i-1) == word2.charAt(j-1))
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = Math.min(dp[i-1][j], Math.min(dp[i][j-1], dp[i-1][j-1])) + 1;
            }
        }
        
        return dp[m][n];
    }