描述
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].
解析
根据题意，可以看出这种求最优的问题可以用 DP 算法来解决，其实直接使用 LCS 的 ASCII 之和作为状态就行了。设 dp[i][j] 为 s1 前 i 个字符与 s2 前 j 个字符得到的 LCS 的 ASCII 和。
转移方程是：
对于 s1[0…i−1] 和 s2[0…j−1] 的 LCS 的 ASCII 的和应该是这样的：
1. 若 s1[i−1]==s2[j−1] ，则 dp[i][j] = dp[i - 1][j - 1] + ord(s1[i - 1])
2. 若不相等，则 s1[i−1], s2[j−1] 选择删除一个，dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
时间复杂度为 O(N^2)，空间复杂度为 O(N^2)。
解答
class Solution(object):
def minimumDeleteSum(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: int
"""
l1, l2 = len(s1), len(s2)
dp = [[0] * (l2 + 1) for _ in range(l1 + 1)]
for i in range(1, l1 + 1):
for j in range(1, l2 + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + ord(s1[i - 1])
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
result = sum(map(ord, s1 + s2)) - dp[-1][-1] * 2
return result
运行结果
Runtime: 596 ms, faster than 73.80% of Python online submissions for Minimum ASCII Delete Sum for Two Strings.
Memory Usage: 12.6 MB, less than 94.12% of Python online submissions for Minimum ASCII Delete Sum for Two Strings.
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