【8】20. Valid Parentheses

20. Valid Parentheses

• Total Accepted: 167254
• Total Submissions: 517330
• Difficulty: Easy
• Contributors: Admin

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.

Solution:

Use stack. When we meet a left parentheses, we push it into the stack. When we meet a right parentheses, we will return false if the stack is empty, otherwise, we will pop the top of the stack to find if it is the corresponding left parentheses.

class Solution {
public:
    bool isValid(string s) {
        stack<char> st;
        for(int i = 0; i < s.size(); i++){
            if(s[i] == '(' || s[i] == '{' || s[i] == '['){
                st.push(s[i]);
            }else if(s[i] == ')' || s[i] == '}' || s[i] == ']'){
                if(st.empty()){
                    return false;
                }else{//只需判断false的情况，最后再pop就可以了
                    if(s[i] == ')' && st.top() != '(') return false;
                    if(s[i] == ']' && st.top() != '[') return false;
                    if(s[i] == '}' && st.top() != '{') return false;
                    st.pop();
                }
            }
        }
        //return true;此时不能直接return true，因为stack里可能会有剩余的左括号
        return st.empty();
    }
};