712. Minimum ASCII Delete Sum for Two Strings

Problem：

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.

给定两个字符串s1,s2，删除其中一些元素使字符串相等，要求删除的字符ASCII码总和最小。

解决思路：动态规划。s1[i]==s2[j]的时候，可以两个字符都保留，增加的ASCII=0。dp[i][j]=dp[i-1][j-1];

　　　　　　　　　　s1[i]!=s2[j]的时候，至少删除其中一个。dp[i][j] = min(dp[i-1][j]+s1[i], dp[i][j-1]+s2[j]);

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