583.两个字符串的删除操作.简单易懂注释版

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length(), len2 = word2.length();
        int[][] dp = new int[len1 + 1][len2 + 1];
        // word1字符串为空时，word2字符全删
        for(int i = 0; i <= len1; i++) dp[i][0] = i;
        // word2字符串为空时，word1字符全删
        for(int i = 0; i <= len2; i++) dp[0][i] = i;

        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                // 如果相等，则等于两个字符串前面字符的最小步数
                if(word1.charAt(i - 1) == word2.charAt(j - 1)){
                    dp[i][j] = dp[i - 1][j - 1];
                }else{
                    // dp[i-1][j-1]+2肯定比前面两种情况大，所以舍弃
                    dp[i][j] = Math.min(dp[i - 1][j] + 1, dp[i][j - 1] + 1);
                }
            }
        }
        return dp[len1][len2];
    }
}