Table of Contents
一、中文版
二、英文版
三、My answer
四、解题报告
给你一个由若干 0 和 1 组成的字符串 s
,请你计算并返回将该字符串分割成两个 非空 子字符串(即 左 子字符串和 右 子字符串)所能获得的最大得分。
「分割字符串的得分」为 左 子字符串中 0 的数量加上 右 子字符串中 1 的数量。
示例 1:
输入:s = "011101" 输出:5 解释: 将字符串 s 划分为两个非空子字符串的可行方案有: 左子字符串 = "0" 且 右子字符串 = "11101",得分 = 1 + 4 = 5 左子字符串 = "01" 且 右子字符串 = "1101",得分 = 1 + 3 = 4 左子字符串 = "011" 且 右子字符串 = "101",得分 = 1 + 2 = 3 左子字符串 = "0111" 且 右子字符串 = "01",得分 = 1 + 1 = 2 左子字符串 = "01110" 且 右子字符串 = "1",得分 = 2 + 1 = 3
示例 2:
输入:s = "00111" 输出:5 解释:当 左子字符串 = "00" 且 右子字符串 = "111" 时,我们得到最大得分 = 2 + 3 = 5
示例 3:
输入:s = "1111" 输出:3
提示:
2 <= s.length <= 500
s
仅由字符 '0'
和 '1'
组成。Given a string s
of zeros and ones, return the maximum score after splitting the string into two non-empty substrings (i.e. left substring and right substring).
The score after splitting a string is the number of zeros in the left substring plus the number of ones in the right substring.
Example 1:
Input: s = "011101" Output: 5 Explanation: All possible ways of splitting s into two non-empty substrings are: left = "0" and right = "11101", score = 1 + 4 = 5 left = "01" and right = "1101", score = 1 + 3 = 4 left = "011" and right = "101", score = 1 + 2 = 3 left = "0111" and right = "01", score = 1 + 1 = 2 left = "01110" and right = "1", score = 2 + 1 = 3
Example 2:
Input: s = "00111" Output: 5 Explanation: When left = "00" and right = "111", we get the maximum score = 2 + 3 = 5
Example 3:
Input: s = "1111" Output: 3
Constraints:
2 <= s.length <= 500
s
consists of characters '0' and '1' only.class Solution:
def maxScore(self, s: str) -> int:
res = 0
for i in range(1,len(s)):
tmp = s[:i].count('0') + s[i:].count('1')
res = max(res, tmp)
return res
因为 s 的长度不会超过 500,所以可以使用暴力遍历方法:
截断每个位置,判断截断前半部分中 0 的个数和截断后半部分中 1 的个数之和。
与 res 比较,选出最大值,赋给 res。