LeetCode Valid Parentheses（020）解法总结Given a string containing

描述

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

An input string is valid if:

Open brackets must be closed by the same type of brackets.
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

思路

使用栈的数据结构，当遇到左括号时入栈，碰到匹配的右括号是出栈。

要注意栈空时的出栈溢出和不匹配的右括号的报错。

class Solution {
    public boolean isValid(String s) {
        LinkedList<Character> stack = new LinkedList<Character>();
        int len = s.length(), count = 0;
        //对字符串进行遍历
        while(count < len){
            if(s.charAt(count) == '('||s.charAt(count) == '['||s.charAt(count) == '{'){
                //左括号入栈
                stack.add(s.charAt(count));
            }else{
                //右括号出栈
                //当栈空/未匹配到时返回false，否则按规则匹配
                if(stack.size() == 0){
                    return false;
                }else if(s.charAt(count) == ')' && stack.getLast() == '('){
                    stack.removeLast(); 
                }else if(s.charAt(count) == ']' && stack.getLast() == '['){
                    stack.removeLast(); 
                }else if(s.charAt(count) == '}' && stack.getLast() == '{'){
                    stack.removeLast(); 
                }else{
                    return false;
                }
            }
            count++;
        }
        //遍历整个输入后栈空即为合法
        if(stack.size() == 0){
            return true;
        }else{
            return false;
        }
    }
}

Runtime: 1 ms, faster than 98.70% of Java online submissions for Valid Parentheses.

Memory Usage: 37.4 MB, less than 5.06% of Java online submissions for Valid Parentheses.

这里使用的是LinkedList作为存储结构，其实java中自带是stack结构的。