[LC] 20. Valid Parentheses

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

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "{[]}"
Output: true

Time: O(N)

Method 1:

class Solution:
    def isValid(self, s: str) -> bool:
        if s is None:
            return False
        pair = {')': '(', '}': '{', ']':'['}
        res = []
        for char in s:
            if char in pair.values():
                res.append(char)
            elif char in pair.keys():
                if res and res[-1] == pair.get(char):
                    res.pop()
                else:
                    return False
            else:
                return False
        return res == []

class Solution {
    public boolean isValid(String s) {
        LinkedList<Character> stack = new LinkedList<>();
        for (char c: s.toCharArray()) {
            if (c == '(' || c == '[' || c == '{') {
                stack.offerFirst(c);
            } else if (c == ']') {
                if (stack.isEmpty() || stack.pollFirst() != '[') {
                    return false;
                }
            } else if (c == ')') {
                if (stack.isEmpty() || stack.pollFirst() != '(') {
                    return false;
                }
            } else if (c == '}') {
                if (stack.isEmpty() || stack.pollFirst() != '{') {
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}

Method 2:

class Solution:
    def isValid(self, s: str) -> bool:
        if s is None:
            return False
        left = ['(', '{', '[']
        right = [')', '}', ']']
        res = []
        for char in s:
            if char in left:
                res.append(char)
            elif char in right:
                if len(res) != 0 and self.match(res[-1], char):
                    res.pop()
                else:
                    return False
        return len(res) == 0
    
    def match(self, l, r):
        return l=='(' and r==')' or l=='{' and r=='}' or l=='[' and r==']'