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题目

Balanced strings are those who have equal quantity of ‘L’ and ‘R’ characters.

Given a balanced string s split it in the maximum amount of balanced strings.

Return the maximum amount of splitted balanced strings.

Example 1:

Input: s = “RLRRLLRLRL”
Output: 4
Explanation: s can be split into “RL”, “RRLL”, “RL”, “RL”, each substring contains same number of ‘L’ and ‘R’.

Example 2:

Input: s = “RLLLLRRRLR”
Output: 3
Explanation: s can be split into “RL”, “LLLRRR”, “LR”, each substring contains same number of ‘L’ and ‘R’.

Example 3:

Input: s = “LLLLRRRR”
Output: 1
Explanation: s can be split into “LLLLRRRR”.

Example 4:

Input: s = “RLRRRLLRLL”
Output: 2
Explanation: s can be split into “RL”, “RRRLLRLL”, since each substring contains an equal number of ‘L’ and ‘R’

Constraints:

1 <= s.length <= 1000
s[i] = 'L' or 'R'

分析

1. 当s中的L和R的个数相同时（不用考虑顺序），则称s是平衡的
2. “最大切分原则”，如“RLRRRLLRLL”，不能切分为RLRRRL****LRLL=3.应该切分为RL、RRRLLRLL=2

初步分析是贪心算法：
将s分成若干个小的字符串r，r中的RL个数相同，每次切分保证r的长度最大。

不理解的部分：如果是贪心，那么例一的“RLRRLLRLRL”应该直接等于1？

不会，看看评论区大神。

评论区解法：
用一个数count记录，如果为R则count++，如果为L则count–，若最终count为0，则平衡。

先从s中拿到一个字母，然后向后匹配，如果平衡，则进行下一轮，如果不平衡，则继续向后匹配。
所以不是贪心算法。

解答

评论区答案(这个慢很多，时间2ms)

class Solution {
    public int balancedStringSplit(String s) {
        int res = 0, cnt = 0;
        for (int i = 0; i < s.length(); ++i) {
            cnt += s.charAt(i) == 'L' ? 1 : -1;
            if (cnt == 0) ++res;
        }
        return res;             
    }    
}

我的答案（这个执行时间0ms）

class Solution {
    public int balancedStringSplit(String s) {
        int tmp=0;
        int res=0;
        for(char c :s.toCharArray()){
            if (c=='R')
                tmp++;
            else
                tmp--;
            if (tmp==0)
                res++;
        }
        return res;
    }
}

将if-else改成三元表达式

class Solution {
    public int balancedStringSplit(String s) {
        int tmp=0;
        int res=0;
        for(char c :s.toCharArray()){
            tmp += c=='R'?1:-1;
            if(tmp==0)
                res++;
        }
        return res;
    }
}