leetcode : edit distance[hard][经典动态规划]

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

状态方程 f[i][j] 表示word1的前i个字符转换成word2的前jg字符所需要的最小次数。

两种情况：  

if(word1.charAt(i - 1) == word2.charAt(j - 1)) {
f[i][j] = Math.min(f[i - 1][j - 1], Math.min(f[i][j - 1] + 1, f[i - 1][j] + 1));
} else {
f[i][j] = Math.min(f[i - 1][j - 1] + 1, Math.min(f[i][j - 1] + 1, f[i - 1][j] + 1));
}

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