Leetcode 712. Minimum ASCII Delete Sum for Two Strings

https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/description/
http://blog.youkuaiyun.com/rrrfff/article/details/7523437
最长公共子序列解析
https://discuss.leetcode.com/topic/107980/c-dp-with-explanation
转移方程的详解来自@zestypanda，非常感谢！

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].

这题就是LCS的变种，删除多余的字符，使得s1和s2相等，只不过需找过程中删除的cost最小

d[i][j]表示s1.substring(0,i)和s2.substring(0,j)的LCS的长度, 那么

if (s1[i-1]==s2[j-1])
    // 相等LCS长度+1
    d[i][j]=d[i-1][j-1]+1;
else
    // 否则从前面找
    d[i][j]=max(d[i-1][j], d[i][j-1]);

d[i][j]表示s1.substring(0,i)和s2.substring(0,j)删除若干个字符后相等的最小cost，那么

if (s1[i-1]==s2[j-1])
    // 不需要删除，cost为0 不增加
    d[i][j]=d[i-1][j-1] + 0;
else
    // 删除s1[i-1]或者s2[j-1]，这样才能让删除后的s1和s2相等
    // 删除的代价就是ASCII值
    d[i][j]=min(d[i-1][j] + s1[i-1], d[i][j-1]+s2[j-1]);

初始化边界，若s1和s2中有空串，那么就得全部删掉才能使得两个空串相等，所以边界就是空串，d[i][0]和d[0][j]，当然d[0][0]表示都是空串，不需要删除，代价为0

d[0][0] = 0;
for (int i = 1; i <= s1.size(); ++i)
    d[i][0] = d[i-1][0] + s1[i-1];
for (int j = 1; j <= s2.size(); ++j)
    d[0][j] = d[0][j-1] + s2[j-1];

代码如下：

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int n = s1.size();
        int m = s2.size();
        int dp[n+1][m+1] = {0};
        for (int i = 1; i <= n; i++) {
            dp[i][0] = dp[i-1][0] + s1[i-1];
        }
        for (int j = 1; j <= m; j++) {
            dp[0][j] = dp[0][j-1] + s2[j-1];
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; 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[n][m];
    }
};