20_Valid-Parentheses

Posted 2020-11-18 xugenpeng

20_Valid-Parentheses

[TOC]

Description

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

Solution

Java solution 1

class Solution {
    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<>();
        for (int i=0; i<s.length(); i++) {
            char c = s.charAt(i);

            if (c == '(' || c == '[' || c == '{') {
                stack.push(c);
            } else {
                if (stack.empty()) {
                    return false;
                }

                char topChar = stack.pop();
                if (c == ')' && topChar != '(') {
                    return false;
                }
                if (c == ']' && topChar != '[') {
                    return false;
                }
                if (c == '}' && topChar != '{') {
                    return false;
                }
            }
        }

        return stack.isEmpty();
    }
}

Runtime:?15 ms

Java solution 2 (Better!)

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

Runtime:?12 ms

Python solution

class Solution:
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        stack = list()
        dict = {'(': ')', '[': ']', '{': '}'}
        for c in s:
            if c in dict.keys():
                stack.append(dict[c])
            elif c in dict.values():
                if len(stack) == 0 or c != stack.pop():
                    return False
            else:
                False
        return len(stack) == 0

Runtime:?40 ms