【LeetCode算法】Valid Parentheses-优快云博客

本文解析了LeetCode第20题的有效括号字符串问题，介绍了两种解决方法：一种是通过递归删除成对的括号，另一种是使用栈来验证括号的有效性，并详细分析了使用栈的方法。

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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:

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：

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;
        }
    }
}