题目

开始以为不是dp 后来看了下题解
dp[i][j] 代表word1截止到第i个字符转换成word2截止到第j个字符需要几步变换

状态方程

if(word1[i] == word2[j]) dp[i][j] = dp[i-1][j-1] //无需操作
else dp[i][j] = Math.min(Math.min(dp[i][j-1],dp[i-1][j]),dp[i-1][j-1])+1;
dp[i][j-1]+1 代表word1截止到第i个转换成word2第j-1个需要的次数然后再插入第j个字符
dp[i-1][j]+1 代表word1截止到第i-1个转换成word2第j个需要的次数然后再删除第i个字符
dp[i-1][j-1] 代表word1截止到第i-1个转换成word2第j-1个需要的次数然后再替换第i个字符变为第j个字符

    public int minDistance(String word1, String word2) {
    
    
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1+1][len2+1];
        for (int i = 1; i < len2 + 1; i++)
            dp[0][i] = dp[0][i-1] + 1;
        for (int j = 1; j < len1 + 1; j++)
            dp[j][0] = dp[j-1][0] + 1;
        for (int i = 1; i < len1 + 1; i++) {
    
    
            for (int j = 1; j < len2 + 1; j++) {
    
    
                if (word1.charAt(i-1) == word2.charAt(j-1))
                    dp[i][j] = dp[i-1][j-1];
                else
                    dp[i][j] = Math.min(Math.min(dp[i][j-1],dp[i-1][j]),dp[i-1][j-1])+1;
            }
        }
        return dp[len1][len2];
    }