原文链接: https://leetcode-cn.com/problems/maximum-score-after-splitting-a-string

英文原文

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
• The string s consists of characters '0' and '1' only.

中文题目

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
• 字符串 s 仅由字符 '0' 和 '1' 组成。

通过代码

高赞题解

代码通俗易懂~
流程全写在做注释上了~~~

class Solution {
    public int maxScore(String s) {
        int res = 0, cnt1 = 0, cnt0 = 0;        //cnt1统计右边1的个数，同理cnt0左边0的个数
        for(int i = 0; i < s.length(); i++){
            cnt1 += s.charAt(i)-'0';            //先统计1的个数
        }                                       //由于左右区域的数至少为1，所以i不能等于len-1
        for(int i = 0; i < s.length()-1; i++){  //点i分为左右两个区域        
            if(s.charAt(i) == '0') cnt0++;      //遇到01就统计，动态更新左右区域01个数
            else cnt1--;
            res = Math.max(res, cnt0+cnt1);
        }
        return res;
    }
}

统计信息

通过次数	提交次数	AC比率
10910	20187	54.0%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言