leetcode 20 Valid Parentheses 括号匹配

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and‘]‘, determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

写了一个0ms 的代码：

// 20150630.cpp : 定义控制台应用程序的入口点。

//

#include "stdafx.h"
#include <iostream>
#include <stack>
#include <string>
using namespace std;

bool isValid(string s)
{
	if (s=="")return false;

	stack<char> Parentheses;
	int size =s.size();

	Parentheses.push(s[0]);


	for(int i = 1;i < size ;++i)
	{

		if(Parentheses.top()==‘(‘&&s[i]==‘)‘||Parentheses.top()==‘[‘&&s[i]==‘]‘||Parentheses.top()==‘{‘&&s[i]==‘}‘)
		{
			Parentheses.pop();
			if (Parentheses.empty()&&(i+1)!=size)
			{
				Parentheses.push(s[i+1]);
				i++;
			}
		}

		else
		{
			Parentheses.push(s[i]);
		}


	}

	if(Parentheses.empty())
	{
		return true;
	}
	else
	{
		return false;
	}
}
int _tmain(int argc, _TCHAR* argv[])
{
	string s = "()[]{}";
	isValid(s);
	return 0;
}

另外一个看着好看点的：

class Solution {
    public:
        bool isValid(string s)
        {
            std::stack<char> openStack;
            for(int i = 0; i < s.length(); i++)
            {
                switch(s[i])
                {
                    case ‘(‘:
                    case ‘{‘:
                    case ‘[‘:
                        openStack.push(s[i]);
                        break;
                    case ‘)‘:
                        if(!openStack.empty() && openStack.top() == ‘(‘ )
                            openStack.pop();
                        else
                            return false;
                        break;
                    case ‘}‘:
                        if(!openStack.empty() && openStack.top() == ‘{‘ )
                            openStack.pop();
                        else
                            return false;
                        break;
                    case ‘]‘:
                        if(!openStack.empty() && openStack.top() == ‘[‘ )
                            openStack.pop();
                        else
                            return false;
                        break;

                    default:
                        return false;
                }
            }

            if(openStack.empty())
                return true;
            else
                return false;
        }
    };

python代码：

class Solution:
    # @return a boolean
    def isValid(self, s):
        stack = []
        dict = {"]":"[", "}":"{", ")":"("}
        for char in s:
            if char in dict.values():
                stack.append(char)
            elif char in dict.keys():
                if stack == [] or dict[char] != stack.pop():
                    return False
            else:
                return False
        return stack == []

</pre><pre class="python" name="code">class Solution:
    # @param s, a string
    # @return a boolean
    def isValid(self, s):
        paren_map = {
            ‘(‘: ‘)‘,
            ‘{‘: ‘}‘,
            ‘[‘: ‘]‘
        }
        stack = []

        for p in s:
            if p in paren_map:
                stack.append(paren_map[p])
            else:
                if not stack or stack.pop() != p:
                    return False

        return not stack

class Solution:
    # @param s, a string
    # @return a boolean
    def isValid(self, s):
        d = {‘(‘:‘)‘, ‘[‘:‘]‘,‘{‘:‘}‘}
        sl = []
        for i in s:
            if i in d:
                sl.append(i)
            else:
                if not sl or d[sl.pop()] != i:
                    return False

        if sl:
            return False
        return True