题目描述：

LeetCode 712. Minimum ASCII Delete Sum for Two Strings

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].

题目大意：

给定字符串s1, s2。删除其中的一些字符使得两字符串相等

删除字符的代价是被删字符ASCII之和，求最小代价。

解题思路：

动态规划（Dynamic Programming）

该题目是编辑距离的变体，可以参考编辑距离的状态转移方程

Python代码：

class Solution(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        a1, a2 = map(ord, s1), map(ord, s2)
        l1, l2 = len(s1), len(s2)
        dp = [0]
        for x in range(l1):
            dp.append(dp[-1] + a1[x])
        for x in range(1, l2 + 1):
            ndp = [dp[0] + a2[x - 1]]
            for y in range(1, l1 + 1):
                if a2[x - 1] == a1[y - 1]: ndp.append(dp[y - 1])
                else: ndp.append(min(dp[y] + a2[x - 1], ndp[y - 1] + a1[y - 1]))
            dp = ndp
        return dp[-1]

 

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