zoukankan      html  css  js  c++  java
  • 【LeetCode & 剑指offer刷题】字符串题11:Valid Parentheses(括号对)

    【LeetCode & 剑指offer 刷题笔记】目录(持续更新中...)

    Valid Parentheses

    Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
    An input string is valid if:
    1. Open brackets must be closed by the same type of brackets.
    2. Open brackets must be closed in the correct order.
    Note that an empty string is also considered valid.
    Example 1:
    Input: "()"
    Output: true
    Example 2:
    Input: "()[]{}"
    Output: true
    Example 3:
    Input: "(]"
    Output: false
    Example 4:
    Input: "([)]"
    Output: false
    Example 5:
    Input: "{[]}"
    Output: true

    C++
     
    /*问题:有效括号对判断
    #include<stack> 在该环境下已经包含
    方法一:
    左括号入栈,右括号消解使栈顶元素出栈,若右括号不能与栈顶元素匹配,则返回false
    */
    class Solution
    {
        public:
            bool isValid(string s)
            {
               
                stack<char> stk;
                for(auto c: s)
                {
                    switch(c)
                    {
                        case '(':
                        case '{':
                        case '[':
                            stk.push(c);break; //左括号入栈
                           
                        case ')':
                            if(stk.empty() || stk.top()!='(') //如果栈顶元素为不匹配的括号时,说明不能构成括号对(如example 4)
                                return false;
                            else
                                stk.pop();break; //右括号时原左括号被消解出栈
                        case '}':
                            if(stk.empty() || stk.top()!='{')
                                return false;
                            else
                                stk.pop();break;
                        case ']':
                            if(stk.empty() || stk.top()!='[')
                                return false;
                            else
                                stk.pop();break;
                    }
                }
               
                return stk.empty(); //消解完后看栈中是否还有元素,如果还有则为false
               
            }
    };
    //方法二
    /*class Solution
    {
        public:
            bool isValid(string s)
            {
                string left = "([{";
                string right = ")]}";
                stack<char> stk;
               
                for(auto c: s)
                {
                    if(left.find(c) != string::npos) //如果为左括号则入栈
                    {
                        stk.push(c);
                    }
                    else //如果为右括号,看栈顶元素是否为左括号
                    {
                        if(stk.empty() || stk.top()!=left[right.find(c)])
                            return false;
                        else//匹配消解出栈
                            stk.pop();
                    }
                }
               
                return stk.empty(); //消解完后看栈中是否还有元素,如果还有则为false
               
            }
    };*/
     
  • 相关阅读:
    搭建strom 的开发环境
    maven 的plugin 的使用
    Maven 的dependency 的 classifier的作用
    Maven中的dependency的scope作用域详解
    Supervisor-进程监控自动重启
    websocket 实战
    vue 监听路由变化
    vux-uploader 图片上传组件
    vue 定义全局函数
    判断对象属性的值是否空,如为空,删除该属性
  • 原文地址:https://www.cnblogs.com/wikiwen/p/10224904.html
Copyright © 2011-2022 走看看