1221. 分割平衡字符串

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

Java

/*
 * @Author: Goog Tech
 * @Date: 2020-08-15 11:46:02
 * @LastEditTime: 2020-08-15 11:58:35
 * @Description: https://leetcode-cn.com/problems/split-a-string-in-balanced-strings/
 * @FilePath: \leetcode-googtech\#1221. Split a String in Balanced Strings\Solution.java
 * @WebSite: https://algorithm.show/
 */

/*
 * @lc app=leetcode.cn id=1221 lang=java
 *
 * [1221] 分割平衡字符串
 */

// @lc code=start
class Solution {

    // 利用栈
    public int balancedStringSplit(String s) {
        int result = 0;
        // 初始化辅助栈
        Stack<Character> stack = new Stack<>();
        // 遍历栈中字符串中的字符
        for(int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);
            // 若栈中无元素或栈顶字符与当前字符一致时,则将当前字符入栈,反之出栈
            if(stack.isEmpty() || ch == stack.peek()) {
                stack.push(ch);
            }else {
                stack.pop();
            }
            // 检查当前栈是否为空,若为空则平衡次数加一
            if(stack.isEmpty()) {
                result++;
            }
        }
        // 返回结果
        return result;
    }
}
// @lc code=end

Python

'''
Author: Goog Tech
Date: 2020-08-15 11:54:14
LastEditTime: 2020-08-15 11:58:43
Description: https://leetcode-cn.com/problems/split-a-string-in-balanced-strings/
FilePath: \leetcode-googtech\#1221. Split a String in Balanced Strings\Solution.py
WebSite: https://algorithm.show/
'''

#
# @lc app=leetcode.cn id=1221 lang=python
#
# [1221] 分割平衡字符串
#

# @lc code=start
class Solution(object):

    # 贪心算法
    def balancedStringSplit(self, s):
        """
        :type s: str
        :rtype: int
        """
        balance, result = 0, 0
        for ch in s:
            if ch == 'L': balance += 1
            elif ch == 'R': balance -= 1
            if balance == 0: result += 1
        return result
# @lc code=end