[LeetCode]712. Minimum ASCII Delete Sum for Two Strings

• descrption
  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: 231Explanation: 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: 403Explanation: 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
  题意是将两个字符串删除个别字符后得到相同的字符串，求得删除字符的ASCII码的值的最小和，那么反向思考也是求得剩下字符的ASCII码的和的最大值，因为字符串中字符的总的ASCII码值的总和是不变的。下面就可以将问题转化为：求得两个字符串中公共子序列的最大的ASCII码值的和，对求最长公共子序列的方法稍加改变就可得到该问题的答案。

求解最长子序列的状态转移方程为:用l[i][j]记录以s1[i]和s2[j]结尾的最长子序列的长度if s1[i] == s2[j]l[i][j] = l[i-1][j-1] + 1;else l[i][j] = max(l[i-1][j], l[i][j-1]);==========================================================求解该问题的状态转移方程为：用dp[i][j]记录以s1[i]和s2[j]结尾的子序列的最大的ASCII码值的总和if s1[i] == s2[j]dp[i][j] = dp[i-1][j-1] + (int)s1[i]; elsedp[i][j] = max(dp[i-1][j], dp[i][j-1]);

• 代码如下

class Solution {public:    int minimumDeleteSum(string s1, string s2) {        int len1 = s1.length();        int len2 = s2.length();        int dp[len1+1][len2+1];        memset(dp, 0, sizeof(dp));        for (int i = 0; i < len1; i++) {            for (int j = 0; j < len2; j++){                if(s1[i] == s2[j]) {                    dp[i+1][j+1] = dp[i][j] + (int)s1[i];                 }                else {                    dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j]);                }            }        }        int all = 0;        for(int i = 0; i < len1; i ++) {            all += (int)s1[i];        }        for(int j = 0; j < len2; j ++) {            all += (int)s2[j];        }        return all - 2*dp[len1][len2];    }};

原题地址

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