Leetcode 20: 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.

这题判断括号对，这种linear 往前看的，需要用到stack记录之前的状态，只有当stack的top和current element能匹配，stack才能pop出element，否则一直往里面加element，最后整个string遍历完，看stack里面是否为空，如果为空，说明完全匹配，否则不能完全匹配。
class Solution {
public:
    bool isValid(string s) {
        //sanity check
        if (s.size() == 0) return true;
        stack<char> buf;
        for(int i = 0; i < s.size(); i++) {
            if (buf.empty()) {
                buf.push(s[i]);
                continue;
            }
            if ((s[i] == ')' && buf.top() == '(') || (s[i] == '}' && buf.top() == '{') ||(s[i] == ']' && buf.top() == '[')) {
                buf.pop();
            }else buf.push(s[i]);
        }
        //to check if buf is empty
        if (buf.empty()) return true;
        else return false;
    }
};
time complexity： O（n)
space complexity: O(1)

updated:

more shorter code, there are only one condition to pop from the stack, when the parentheses matches, in other conditions, just push into the stack
class Solution {
public:
    bool isValid(string s) {
        //sanity check
        if(s.size() == 0) return true;
        stack<char> buf;
        for (int i = 0; i < s.size(); i++) {
                if (!buf.empty() && (buf.top() == '{' && s[i] == '}' || (buf.top() == '[' && s[i] == ']') || (buf.top() == '(' && s[i] == ')')))
                buf.pop();
                else buf.push(s[i]);
        }
        if (buf.empty()) return true;
        else return false;
    }
};