[LeetCode]Valid Parentheses

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.

思考：{ [ ( 入栈，) ] } 出栈匹配。

class Solution {
public:
    bool isValid(string s) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int len=s.size();
		int i;
		bool ans=true;
		stack<char> p;
		char temp;
		for(i=0;i<len;i++)
		{
			if(s[i]=='('||s[i]=='['||s[i]=='{')
				p.push(s[i]);
			else if(s[i]==')')
			{
				if(p.empty()) return false;
				else
				{
					temp=p.top();
					if(temp!='(') return false;
					p.pop();
				}
			}
			else if(s[i]==']')
			{
				if(p.empty()) return false;
				else
				{
					temp=p.top();
					if(temp!='[') return false;
					p.pop();
				}
			}
			else if(s[i]=='}')
			{
				if(p.empty()) return false;
				else
				{
					temp=p.top();
					if(temp!='{') return false;
					p.pop();
				}
			}
		}
		if(p.size()>0) return false;
		return ans;
    }
};

2014-03-27 16:13:30

class Solution {
public:
    bool isValid(string s) {
        int len=s.size();
        if(len==0) return true;
        stack<char> temp;
        int i=0;
        char ch;
        while(i<len)
        {
            if(s[i]=='('||s[i]=='['||s[i]=='{') temp.push(s[i]);
            else
            {
                if(temp.empty()) return false;
                ch=temp.top();
                if(s[i]==')'&&ch!='(') return false;
                if(s[i]==']'&&ch!='[') return false;
                if(s[i]=='}'&&ch!='{') return false;
                temp.pop();
            }
            i++;
        }
        if(!temp.empty()) return false;
        else return true;
    }
};