[LeetCode] DP之 Minimum ASCII Delete Sum for two Strings

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, which are higher.

Note:

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

分析

这题和课本上的编辑距离很类似，都是像列表格一样的dp思想，但是区别在于将考虑的cost变成了删除的字符的ASCII之和，其实这也是一种有点像贪心的做法，每次选取的次数都是最小的，最优子结构。
关键思想在于考虑一个minCost[i][j]，这个的值是子串s1.substring(0, i-1), s2.substring(0,j-1)的minimum ASCII delete sum。因此，对于minCost[i+1][j+1] 只会有两种情况，要么s1[i] == s2[j]，此时minCost[i+1][j+1] = minCost[i][j]。 而当不等时，应该是删除s1[i]和删除s2[j]二选一，那么就是min(minCost[i][j+1] + s1[i] , minCost[i+1][j] +s2[j]) 之一。

时间复杂度分析

时间复杂度很简单，假设m和n分别是s1和s2的size，那么就是O(mn)

代码

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int minCost[s1.size()+1][s2.size()+1];
        minCost[0][0] = 0;
        for (int i = 1; i <= s1.size(); i++) {
            minCost[i][0] = minCost[i-1][0] + s1[i-1];
            for (int j = 1; j <= s2.size(); j++) {
                minCost[0][j] = minCost[0][j-1] +s2[j-1];
                if (s1[i-1] != s2[j-1]) minCost[i][j] = min(minCost[i-1][j]+ s1[i-1], minCost[i][j-1] + s2[j-1]);
                else minCost[i][j] = minCost[i-1][j-1];
            }
        }
        return minCost[s1.size()][s2.size()];
     }
};