zoukankan      html  css  js  c++  java
  • 【leetcode】1190. Reverse Substrings Between Each Pair of Parentheses

    题目如下:

    Given a string s that consists of lower case English letters and brackets. 

    Reverse the strings in each pair of matching parentheses, starting from the innermost one.

    Your result should not contain any bracket.

    Example 1:

    Input: s = "(abcd)"
    Output: "dcba"
    

    Example 2:

    Input: s = "(u(love)i)"
    Output: "iloveu"
    

    Example 3:

    Input: s = "(ed(et(oc))el)"
    Output: "leetcode"
    

    Example 4:

    Input: s = "a(bcdefghijkl(mno)p)q"
    Output: "apmnolkjihgfedcbq"
    

    Constraints:

    • 0 <= s.length <= 2000
    • s only contains lower case English characters and parentheses.
    • It's guaranteed that all parentheses are balanced.

    解题思路:本题和leetcode之前出现过的四则运算的题目类似。从头开始遍历s,不是'('的字符直接入栈,如果遇到')',找出栈中最靠近栈顶的'(',逆置从'('到栈顶的所有元素,同时删除'(',直到s遍历完成为止。

    代码如下:

    class Solution(object):
        def reverseParentheses(self, s):
            """
            :type s: str
            :rtype: str
            """
            left_inx = []
            stack = []
            for i in s:
                if i == '(':
                    stack.append(i)
                    left_inx.append(len(stack)-1)
                elif i == ')':
                    inx = left_inx.pop(-1)
                    sub = stack[inx + 1:]
                    sub.reverse()
                    stack = stack[:inx] + sub
                else:
                    stack.append(i)
            return ''.join(stack)
  • 相关阅读:
    第十三次ScrumMeeting会议
    第十二次ScrumMeeting博客
    第十一次ScrumMeeting会议
    Alpha阶段事后分析
    Alpha阶段展示
    Alpha阶段发布说明
    团队项目-第十次scrum 会议
    团队项目-第五次Scrum 会议
    # 团队项目测评博客
    # Beta冲刺前准备
  • 原文地址:https://www.cnblogs.com/seyjs/p/11532014.html
Copyright © 2011-2022 走看看