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].

動態規劃的經典問題，計算兩個字串之間的編輯距離，只不過這裡變成了字元的ASCII碼值距離，關鍵的狀態轉移方程如下, 時間複雜度O(n^2)：

//若相等編輯距離不變
if (s1[i-1] == s2[j-1]) {
    dp[i][j] = dp[i - 1][j - 1];
} else {
//若不等則有兩種情況，取最優的那種情況
    dp[i][j] =min(dp[i][j-1
 
]+s2[j-1], dp[i - 1][j ]+s1[i-1]);
}

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        vector<int> t(s2.size()+1);
        vector<vector<int>> dp;
        for (int i = 0; i < s1.size()+1; i++) {
            dp.push_back(t);
        }
        for (int
 
 i = 0; i < s1.size(); i++) {
            for (int j = 0; j < s2.size(); j++) {
                dp[i][j] = INT_MAX;
            }
        }
        dp[0][0] = 0;
        for (int i = 1; i <= s2.size(); i++) {
            dp[0][i] = dp[0][i-1]+s2[i-1];
        }
        for (int i = 1; i <= s1.size(); i++) {
            dp[i][0] =dp[i-1][0]+s1[i-1];
        }
        for (int i = 1; i <= s1.size(); i++) {
            for (int j = 1; j <= s2.size(); 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[s1.size()][s2.size()];
    }

};