leetcode -- 712. Minimum ASCII Delete Sum for Two Strings

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].

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/minimum-ascii-delete-sum-for-two-strings
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

思路：

1. dp[i][j]表示字符串s1前i个字符，字符串s2前j个字符达到完全一致需要删除的最少ASCII和

2. 边界条件：dp[0][0] = 0

3. dp[i][0] = dp[i-1][0] + s1[i-1]：字符串s2为空时，删除的最少ASCII为s1全部ASCII码值

dp[0][j] = dp[0][j-1]+s2[j-1]：字符串s1为空时，删除的最少ASCII为s2全部ASCII码值

3. 当s[i] == s[j]，dp[i][j] = dp[i-1][j-1]，相等时，s[i] == s[j]时，不需要删除，则为前i-1个和j-1个的dp值

当s[i] != s[j]，dp[i][j] = min(dp[i-1]j]+s1[i-1]，dp[i][j-1]+s2[j-1])，不相等时，要么删除si，要么删除sj，取最小值

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int len1 = s1.length();
        int len2 = s2.length();
        int dp[len1+1][len2+1];
        dp[0][0] = 0;
        for(int i = 1; i <= len1; i++){
            dp[i][0] = dp[i-1][0] + s1[i-1];
        }
        for(int j = 1; j <= len2; j++){
            dp[0][j] = dp[0][j-1] + s2[j-1];
        }
        for(int i = 1; i <= len1; i++){
            for(int j = 1; j <= len2; j++){
                if(s1[i-1] == s2[j-1]){
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    dp[i][j] = min(dp[i][j-1]+s2[j-1], dp[i-1][j]+s1[i-1]);
                }
            }
        }
        return dp[len1][len2];
    }
};

作者：hestyle 
来源：CSDN 
原文：https://blog.csdn.net/qq_41855420/article/details/89576652 
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