动态规划--两字字符串的编辑距离

1、题目

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, whic

2、解答：dp[i][j] = dp[i-1][j-1] (if s1[i] = s2[j]) + min{ dp[i-1][j] + s1[i-1]（删除s1的一个字符）+ dp[i][j-1] + s2[j-1](删除s2的一个字符) }

3、代码：

C++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] + 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] + s1[i-1],dp[i][j-1]+s2[j-1]);
        
        return dp[m][n];
    }
};

python代码

class Solution:
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        m,n = len(s1),len(s2)
        
        path = [[0 for i in range(n+1)] for j in range(m+1)]
        
        for i in range(1,m+1):
            path[i][0] = path[i-1][0] + ord(s1[i-1])
        
        for j in range(1,n+1):
            path[0][j] = path[0][j-1] + ord(s2[j-1])
            
        for i in range(1,m+1):
            for j in range(1,n+1):
                if s1[i-1] == s2[j-1]:
                    path[i][j] = path[i-1][j-1]
                else:
                    path[i][j] = min(path[i-1][j] + ord(s1[i-1]), path[i][j-1] + ord(s2[j-1]))
                    
        return path[-1][-1]