1221. Split a String in Balanced Strings

2021/9/7 23:10:33

本文主要是介绍1221. Split a String in Balanced Strings,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

题目:

Balanced strings are those that have an 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 split 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] is either 'L' or 'R'.
  • s is a balanced string.

思路:

easy题,当一个字符串内L和R的数量一样多时即为平衡字符串。基础思路是遍历字符串,用三个计数器,如果遇到L则 l加一;如果遇到R则 r加一;当l和r相等时,说明前面出现了一个平衡字符串,则count加一,然后把l和r清空,直到遍历完字符串,返回count即可。

代码:

class Solution {
public:
    int balancedStringSplit(string s) {
        int r = 0, l = 0, count = 0;
        for (auto a : s) {
            if (a == 'R')
                r++;
            else if (a == 'L')
                l++;
            if (l != 0 && l == r) {
                count++;
                l = r = 0;
            }
        }
        return count;
    }
};



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