标签:col tco har 二分 长度 equals wan nta nat
题目如下:
You are given two strings
s
andt
of the same length. You want to changes
tot
. Changing thei
-th character ofs
toi
-th character oft
costs|s[i] - t[i]|
that is, the absolute difference between the ASCII values of the characters.You are also given an integer
maxCost
.Return the maximum length of a substring of
s
that can be changed to be the same as the corresponding substring oft
with a cost less than or equal tomaxCost
.If there is no substring from
s
that can be changed to its corresponding substring fromt
, return0
.Example 1:
Input: s = "abcd", t = "bcdf", maxCost = 3 Output: 3 Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.Example 2:
Input: s = "abcd", t = "cdef", maxCost = 3 Output: 1 Explanation: Each character in s costs 2 to change to charactor int, so the maximum length is 1.
Example 3:
Input: s = "abcd", t = "acde", maxCost = 0 Output: 1 Explanation: You can‘t make any change, so the maximum length is 1.Constraints:
1 <= s.length, t.length <= 10^5
0 <= maxCost <= 10^6
s
andt
only contain lower case English letters.
解题思路:本题包了一层壳,去掉外表后题目是给定一个正整数组成的数组,求出最长的一段子数组的长度,要求子数组的和不大于cost。解题方法也不难,记per_cost[i]为abs(s[i] - t[i])的值,cost[i]为sum(per_cost[0:i])的值。对于任意一个下标i,很容易通过二分查找的方法找出cost中另外一个下标j,使得cost[i:j] <= cost。
代码如下:
class Solution(object): def equalSubstring(self, s, t, maxCost): """ :type s: str :type t: str :type maxCost: int :rtype: int """ cost = [] amount = 0 per_cost = [] for cs,ct in zip(s,t): amount += abs(ord(cs) - ord(ct)) cost.append(amount) per_cost.append(abs(ord(cs) - ord(ct))) #cost.sort() #print cost #print per_cost import bisect res = -float(‘inf‘) for i in range(len(cost)): inx = bisect.bisect_right(cost,cost[i] + maxCost - per_cost[i]) res = max(res,inx - i) return res
【leetcode】1208. Get Equal Substrings Within Budget
标签:col tco har 二分 长度 equals wan nta nat
原文地址:https://www.cnblogs.com/seyjs/p/11616703.html