Minimum ASCII Delete Sum for Two Strings

题目描述：
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.

分析：这道题本质上和DP常见题目编辑距离是一样的题目，
用一个二维数组dis,其中dis[i][j]表示S1前i位与S2前j位的最小删除和，
显然，当是S1第i+1位等于S2第j+1位时，dis[i+1][j+1]=dis[i][j],当不相等时,dis[i+1][j+1]为”dis[i][j+1]+S1[j]”与“dis[i+1][j]+S2[i]”的较小值。最终dis[S1.length()][S2.length()]就是所求。

代码如下：

   int minimumDeleteSum(string s1, string s2) {
        int len1=s1.length();
    int len2=s2.length();
    vector<vector<int> >dis(len1+1,vector<int>(len2+1,0));
    for(int i=1;i<=len2;i++)
    dis[0][i]=dis[0][i-1]+s2[i-1];
    for(int j=1;j<=len1;j++)
    dis[j][0]=dis[j-1][0]+s1[j-1];
    for(int i=1;i<=len1;i++)
    for(int j=1;j<=len2;j++){
        if(s1[i-1]==s2[j-1])
        dis[i][j]=dis[i-1][j-1];
        else
        dis[i][j]=min(dis[i][j-1]+s2[j-1],dis[i-1][j]+s1[i-1]);
       }
       return dis[len1][len2];
    }