20. Valid Parentheses - Easy

标签：length   nbsp   col   style   als   false   ber   put   rac   

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

use stack. for each character in input string:

(1) if current char is a open bracket, push it into the stack;

(2) if it‘s a close bracket: 

　　(2.1) if stack is empty, return false directly; 

　　(2.2) if current char does not match the top element of the stack, return false. (remember to pop the top element if matches)

(3) finally, if stack is empty return true, else return false

time = O(n), space = O(n)

class Solution {
    public boolean isValid(String s) {
        LinkedList<Character> stack = new LinkedList<>();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if(c == ‘(‘ || c == ‘{‘ || c == ‘[‘) {
                stack.push(c);
            } else {
                if(stack.isEmpty()) {
                    return false;
                } else if(c == ‘)‘ && stack.pop() != ‘(‘) {
                    return false;
                } else if(c == ‘}‘ && stack.pop() != ‘{‘) {
                    return false;
                } else if(c == ‘]‘ && stack.pop() != ‘[‘) {
                    return false;
                }
            }
        }
        return stack.isEmpty();
    }
}

20. Valid Parentheses - Easy

标签：length   nbsp   col   style   als   false   ber   put   rac   

原文地址：https://www.cnblogs.com/fatttcat/p/11132773.html