1.问题

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

Given a balanced string s, split it into some number of substrings such that:

Each substring is balanced.
Return the maximum number of balanced strings you can obtain.
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 = “RLRRRLLRLL”
Output: 2
Explanation: s can be split into “RL”, “RRRLLRLL”, each substring contains same number of ‘L’ and ‘R’.
Note that s cannot be split into “RL”, “RR”, “RL”, “LR”, “LL”, because the 2nd and 5th substrings are not balanced.

Example 3:

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

Constraints:

2 <= s.length <= 1000
s[i] is either ‘L’ or ‘R’.
s is a balanced string.

2.解题思路

方法1：

解决这个问题的方法是循环遍历字符串，并在遇到时不断增加 L 和 R 的计数。
1.定义L 和 R 的计数的变量，以及返回字符串个数的变量sum
2.s转为char[] 方便遍历数组
3.如果字符等于”R“，R的计数加1，否则L的计数加1
4.R的计数=L的计数,返回字符串个数的变量sum加1
5.返回sum

方法2：

1.定义result，sum初始值为0
2.for循环遍历，如果s的元素为L，sum为1；如果为R，为-1
3.sum中R和L的个数和为0，result加一
4.返回值result

3.代码

代码1：

class Solution {
    public int balancedStringSplit(String s) {
        int countR = 0;//1.定义L 和 R 的计数的变量，以及返回字符串个数的变量sum
        int countL =0;
        int sum=0;
        char[] d = s.toCharArray();//2.s转为char[] 方便遍历数组
        for(int i =0;i<d.length;i++){
            if(d[i]=='R'){
    //3.如果字符等于”R“，R的计数加1，否则L的计数加1
                countR++;
            }else{
                countL++;
            }
            if(countR == countL){
    //4.R的计数=L的计数,返回字符串个数的变量sum加1，
                sum++;
            }
        }
        return sum;//5.返回sum
    }
}

代码2：

class Solution {
    public int balancedStringSplit(String s) {
        int result = 0, sum = 0;//1.定义result，sum初始值为0
        for (int i = 0; i < s.length(); i++) {
    //2.for循环遍历，如果s的元素为L，sum为1；如果为R，为-1
            sum += s.charAt(i) == 'L' ? 1 : -1;
            if (sum == 0) result++;//3.sum中R和L的个数和为0，result加一
        }
        return result;//4.返回值result
    }
}

以下代码和代码2的解题思路相同

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