1. 題目鏈接：

https://leetcode.com/problems/split-a-string-in-balanced-strings/

Balanced strings are those who have equal quantity of 'L' and 'R' characters.
Given a balanced string 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".

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

2. 題目關(guān)鍵詞

• 難度等級：easy
• 關(guān)鍵詞：
• 語言： C

3. 解題思路

• 第一種：使用兩個計數(shù)器

遍歷原字符串，找到字符'L'---> LNum++; 'R'---> Rnum++;
如果LNum == Rnum; balanceNum++，并且將LNum 和 Rnum置為0拘哨。

// 題目已要求給的是一個平衡字符串
int balancedStringSplit(char * s){
    int RNum = 0;
    int LNum = 0;
    int balanceNum = 0;
    
    for (int i = 0; i < strlen(s); i++) {
        if (s[i] == 'L') {
            RNum++;
        } else if (s[i] == 'R') {
            LNum++;
        }
        
        if ((RNum == LNum) && RNum != 0) {
            balanceNum++;
        }
    }

    return balanceNum;
}

• 思路2：使用一個計數(shù)器
  遍歷字符串時，遇到R炬丸，num++瞪醋；遇到L時，num--绍妨； if num == 0狂窑， balanceNum++;

// 題目已要求給的是一個平衡字符串
int balancedStringSplit(char * s){
    int num = 0;
    int balanceNum = 0;
    
    for (int i = 0; i < strlen(s); i++) {
        if (s[i] == 'L') {
            num++;
        } else if (s[i] == 'R') {
            num--;
        }
        
        if (num == 0) {
            balanceNum++;
        }
    }
    
    return balanceNum;
}