[LeetCode20]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.

Analysis:

The idea is to use stack to store and match () {} [],

when meet ( { [ push it into stack, meet ) } ] compare and pop stack

Java:

public boolean isValid(String s) {
        int len = s.length();
        if(len<=1||len%2==1) return false;
        Stack<Character> pare = new Stack<>();
        for(int i=0;i<len;i++){
        	if(s.charAt(i)=='('||s.charAt(i)=='['||s.charAt(i)=='{'){
        		pare.push(s.charAt(i));
        	}
        	else {
				if(pare.empty())
					return false;
				else {
					char top = pare.peek();
					if((s.charAt(i)==')'&&top=='(')||(s.charAt(i)==']'&&top=='[')||(s.charAt(i)=='}'&&top=='{'))
						pare.pop();
				}
			}
        }
        if(pare.empty()) return true;
        else return false;
    }
c++

bool isValid(string s) {
        if(s.size()%2 == 1 || s.size() == 0)
        return false;
    vector<char> sta;
    sta.push_back(s[0]);
    for(int i=1; i< s.size(); i++){
        if(s[i] == '(' || s[i]=='[' || s[i]=='{'){
            sta.push_back(s[i]);
            continue;
           }
        char current = sta.back();
        if(s[i] == ')' && current != '(' )
            return false;
        if(s[i] == ']' && current != '[')
            return false;
        if(s[i] == '}' && current != '{')
            return false;
        sta.pop_back();

    }
    if(sta.size() != 0 ) return false;
    return true;
    }



你可能感兴趣的:(LeetCode,stack,Parentheses)