712. Minimum ASCII Delete Sum for Two Strings-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_36718317/article/details/79848127

本文介绍了一种算法，用于计算两个字符串通过删除某些字符变为相同字符串时，被删除字符的ASCII值之和最小值。该算法使用动态规划方法，通过构建二维数组dp来记录不同状态下的最优解。

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Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal.

Example 1:

Input: s1 = "sea", s2 = "eat"
Output: 231
Explanation: Deleting "s" from "sea" adds the ASCII value of "s" (115) to the sum.
Deleting "t" from "eat" adds 116 to the sum.
At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.

Example 2:

Input: s1 = "delete", s2 = "leet"
Output: 403
Explanation: Deleting "dee" from "delete" to turn the string into "let",
adds 100[d]+101[e]+101[e] to the sum.  Deleting "e" from "leet" adds 101[e] to the sum.
At the end, both strings are equal to "let", and the answer is 100+101+101+101 = 403.
If instead we turned both strings into "lee" or "eet", we would get answers of 433 or 417, which are higher.

Note:

0 < s1.length, s2.length <= 1000.All elements of each string will have an ASCII value in [97, 122].

题意：

给出两个字符串，删除两个字符串中的一些字符，求最小的被删除字符ASCII值之和。

思路：

与编辑距离很相似。设置数组dp,dp[i][j]表示s1前i个字符和s2前j个字符使得删除之后相等的那些字符的ASCII值之和。那么可以得到如果s1[i-1]=s2[j-1],那么dp[i][j]=dp[i-1][j-1]，否则就是min(dp[i-1][j]+s1[i-1],dp[i][j-1]+s2[j-1])。

注意初始化的问题，以及谨防数组越界。

代码：

class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        int[][] dp = new int[1+s1.length()][1+s2.length()];  
        for(int i=1;i<=s1.length();i++)
            dp[i][0]=dp[i-1][0]+s1.codePointAt(i-1);
        for(int i=1;i<=s2.length();i++)
            dp[0][i]=dp[0][i-1]+s2.codePointAt(i-1);
        for(int i=1;i<=s1.length();i++)
            for(int j=1;j<=s2.length();j++)
            {
                if(s1.charAt(i-1)==s2.charAt(j-1))
                    dp[i][j]=dp[i-1][j-1];
                else
                    dp[i][j]=Math.min(dp[i][j-1]+s2.codePointAt(j-1),dp[i-1][j]+s1.codePointAt(i-1));
            }
        return dp[s1.length()][s2.length()];
    }
}