712. Minimum ASCII Delete Sum for Two Strings

问题描述：

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].

解题思路：

这道题可以用动态规划来解，可以参考edited distances

dp[i][j]： 将s1的前i个字符构成的子串转换成s2前j个字符串构成的子串删掉的最小的ascii码

初始化：

n = s2.size(), m = s1.size();

dp[0][0] = 0;

dp[0][j] (j: 1->n) = dp[0][j-1] + s2[j-1]

dp[i][0] (i: 1->m) = dp[0][i-1] + s1[i-1]

状态转移方程：

dp[i][j] = dp[i-1][j-1]    (s1[i-1] == s2[j-1])

dp[i][j] = min(dp[i-1][j] + s1[i-1], dp[i][j-1] + s2[j-1])

 

dp[i-1][j] + s1[i-1] : 代表删掉当前s1的字母，上一个子问题就是s1前i-1个字母组成的子串变成s2前j个字母组成的子串的最小的ascii码之和

dp[i][j-1] + s2[j-1]：代表删掉当前s2的字母，上一个子问题就是s1前i个字母组成的子串变成s2前j-1个字母组成的子串的最小的ascii码之和

 

代码：

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