动态规划--两字字符串的编辑距离

- 日理万妓 2022-05-23 09:14 193阅读 0赞

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]