712. Minimum ASCII Delete Sum for Two Strings（python+cpp）

题目：

Given two strings s1, s2, find the lowest ASCII sum of deletedcharacters 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 stringsare 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].

解释：
动态规划
dp[i][j]表示s1的前i个字母和s2的前j个字母变成一样所需要的最小ASCII和，则一共有三种情况
1.dp[i-1][j]+s1[i],从dp[i-1][j]到dp[i][j]是多考虑了s1的一个字符，但是s2的字符数没有变，所以想要相同的话必须删除s1[i]，考虑ASCII的话就是加上s1[i]的ASCII
2.dp[i][j-1]+s2[j],删除s2[j]
3.dp[i-1][j-1]+a ,如果s1[i]==s2[j],则a=0,如果s1[i]!=s2[j],则a=s1[i]+s2[j]
4.在上述三种情况中找最小值
5.这种方法在python中超时了，反正速度特别慢，还是用下面的方法吧：
6.最长公共子序列解法：所有的减去2*相等的字符的ASCII的最大值(即最长公共子序列LCS)，等于题目所需的最小值
最长公共子序列只能求长度，如何求具体的元素？
答：求lcs（subsequence）的时候不是+1而是+ord(s[i]) 最后返回的是dp[m][n]而不是maxLen。(lcsubarray和lcsubsequence的写法略有不同)
python代码:

class Solution(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        def lcsubsequence(s1,s2):
            m=len(s1)
            n=len(s2)
            dp=[[0]*(n+1) for _ in xrange(m+1)]
            for i in xrange(1,m+1):
                for j in xrange(1,n+1):
                    if s1[i-1]==s2[j-1]:
                        dp[i][j]=dp[i-1][j-1]+ord(s1[i-1])
                    else:
                        dp[i][j]=max(dp[i-1][j],dp[i][j-1])
            return dp[m][n]
        m=len(s1)
        n=len(s2)
        if m==0 and n==0:
            return 0
        if m==0:
            return sum(map(ord,s2))
        if n==0:
            return sum(map(ord,s1))
        return sum(map(ord,s1+s2))-2*lcsubsequence(s1,s2)

c++代码：

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int m=s1.size(),n=s2.size();
        if (m==0 && n==0)
            return 0;
        if (m==0)
            return accumulate(s2.begin(),s2.end(),0);
        if (n==0)
            return accumulate(s1.begin(),s1.end(),0);
        return accumulate(s1.begin(),s1.end(),0)+accumulate(s2.begin(),s2.end(),0)-2*subsequence(s1,s2);
    };
    int subsequence(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++)
        {
            for (int j=1;j<=n;j++)
            {
                if(s1[i-1]==s2[j-1])
                    dp[i][j]=dp[i-1][j-1]+s1[i-1];
                else
                {
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        return dp[m][n];
    }
};

总结：
这道题和
718. Maximum Length of Repeated Subarray（python+cpp）以及 583. Delete Operation for Two Strings（python+cpp）一起学习，718是求lcsubarray，583是利用了lcsubsequence。
其实对于m和n是否为0的判断是不必要的，因为题目已经限定了字符串不为空，但是python中删掉判断居然变慢了很多，c++倒是不变，为什么？？？不过为了以后lcs的应用，还是把判断加上吧…

712. Minimum ASCII Delete Sum for Two Strings（python+cpp）

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