class Solution {
    //动态规划 
    // dp[i][j] 意味着将word1的前i个字符转化为word2的前j个字符的最少操作数
    // dp[i][0] = i; dp[0][j] = j;
    //递推公式：if (s.charAt(i) == s.charAt(j)) dp[i][j] == dp[i - 1][j - 1]
    // else dp[i][j] = Math.max(dp[i - 1][j -1] + 1, dp[i-1][j] + 1, dp[i][j-1] + 1);
    // 对应替换的
    public int minDistance(String word1, String word2) {
        int n = word1.length();
        int m = word2.length();
        int[][] dp = new int[n + 1][m + 1];
        for (int i = 0; i <= n; i++) {
            dp[i][0] = i;
        }    
        for (int j = 0; j <= m; j++) {
            dp[0][j] = j;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; 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-1] + 1 , Math.min(dp[i-1][j] + 1, dp[i][j-1] + 1));
                }
            }
        }
        return dp[n][m];
    
    }
}