LeetCode算法Valid Parentheses

Posted 2020-11-06 嘉禾世兴

LeetCode第20题

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

思路：

本来我的想法是不管()，[]，{}，都是在一起的，我一对一对的删掉，最后删空了，就符合要求

代码

class Solution {
    public boolean isValid(String s) {
        if(s.length()>Integer.MAX_VALUE){
            return true;
        }
        for(int i = 0;i<3;i++){
            for(int j = 0;j<s.length();j++){
                s = s.replace("()","");
                System.out.println(s);
                s = s.replace("[]","");
                System.out.println(s);
                s = s.replace("{}","");
                System.out.println(s);
            }
        }
        if("".equals(s)){
            return true;
        }else{
            return false;
        }
    }
}

结果

搞这么多符号，这不故意整我吗

百度了下，都说用栈，代码如下

class Solution {
    public boolean isValid(String s) {
        Stack<String> stack = new Stack<String>();
        for (int i = 0; i < s.length(); i++) {
            char candidate = s.charAt(i);
            if (candidate == \'{\' || candidate == \'[\' || candidate == \'(\') {
                stack.push(candidate + "");
            } else {
                if (stack.isEmpty()) {
                    return false;
                }
                if ((candidate == \'}\' && stack.peek().equals("{")) ||
                        (candidate == \']\' && stack.peek().equals("[")) ||
                        (candidate == \')\' && stack.peek().equals("("))) {
                    stack.pop();
                } else {
                    return false;
                }
            }
        }
        if (stack.isEmpty()) {
            return true;
        } else {
            return false;
        }
    }
}

就是利用后进先出的原理，其实跟我的思路差不多，但是性能要好很多，哈哈