Leetcode-Minimum ASCII Delete Sum for Two Strings

Problem

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, which are higher.

Note:

• 0 < s1.length, s2.length <= 1000.
• All elements of each string will have an ASCII value in [97, 122].

具体详见Leetcode

Analysis

类似的题目

这个问题和两道题有点像，第一道是最长公共子串问题，第二道是字符串的Edit distance。此题又有点像最长公共子序列，因为是可以不连续的序列，不像子串问题一样一定需要连续的。Edit Distance 有三种情况，第一种是替换，第二种是增加，第三种是删减。而此题只有删减一种情况。所以此题可以使用动态规划的方法。

算法

设matrix[i][j]为s1.substr(0,i)和s2.substr(0,j)有相同子序列的字符的ASCII码值的总和。

则matrix[i][j]=max(matrix[i−1][j−1]+same(i,j)∗ASCII(s1[i−1]),matrix[i−1][j],matrix[i][j−1]);

最后的答案：ASCII(s1+s2)-2*matrix[s1.size()][s2.size()]

Complexity

时间复杂度：O(n2)
空间复杂度：O(n2)

code

class Solution {public:    int minimumDeleteSum(string s1, string s2) {        int ** matrix = new int*[s1.size()+1];        for (int i = 0; i <= s1.size(); i++)            matrix[i] = new int[s2.size()+1];        for (int i = 0; i <= s1.size(); i++) {            matrix[i][0] = 0;        }        for (int i = 1; i <= s2.size(); i++) {            matrix[0][i] = 0;        }        int sum = 0;        for (int i = 0; i < s1.size(); i++) {            sum += (s1[i] - 'a'+97);        }        for (int i = 0; i < s2.size(); i++) {            sum += (s2[i] - 'a'+97);        }        for (int i = 1; i <= s1.size(); i++) {            for (int j = 1; j <= s2.size(); j++) {                int sim = (s1[i-1] == s2[j-1]) ? (s1[i-1]-'a'+97) : 0;                matrix[i][j] = max(matrix[i-1][j-1]+sim,matrix[i-1][j]);                matrix[i][j] = max(matrix[i][j-1],matrix[i][j]);            }        }        int result = sum - 2*matrix[s1.size()][s2.size()];        for (int i = 0; i <= s1.size(); i++)            delete []matrix[i];        delete matrix;        return result;    }};