【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
           
        }
};*/
 

 

posted @ 2019-01-05 16:04  wikiwen  阅读(396)  评论(0编辑  收藏