Leetcode 20 Valid Parentheses

最新推荐文章于 2022-01-09 05:17:54 发布

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最新推荐文章于 2022-01-09 05:17:54 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/83096124

本文详细解析了如何通过栈数据结构实现括号匹配算法，包括圆括号、方括号和花括号的正确配对验证。提供了两种实现方法，一种使用栈来逐个检查括号，另一种使用数组和自定义函数进行优化。文章适合初学者理解和实践字符串匹配算法。

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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

这个题是简单的字符串匹配

1）

class Solution {
    public boolean isValid(String s) {
        if(s == null || s.length() < 1){
            return true;
        }else if(s.length() == 1){
            return false;
        }
        Stack<Character> stack = new Stack<>();
        for(int i = 0 ; i < s.length();i++){
            if(s.charAt(i) != ')' && s.charAt(i) != ']' && s.charAt(i) != '}'){
                stack.push(s.charAt(i));
            }else{
                if(!stack.isEmpty() && stack.peek() == '(' && s.charAt(i) == ')'){
                    stack.pop();
                }else if(!stack.isEmpty() && stack.peek() == '[' && s.charAt(i) == ']'){
                    stack.pop();
                }else if(!stack.isEmpty() && stack.peek() == '{' && s.charAt(i) == '}'){
                    stack.pop();
                }else{
                    return false;
                }
            }
        }
        if(!stack.isEmpty()){
            return false;
        }else{
            return true;
        }
    }
}

2）

class Solution {
    public boolean isValid(String s) {
        int top = -1;
        char[] cs = s.toCharArray();
        for(int i=0;i<s.length();i++){
            if(top < 0 || !ism(cs[top], cs[i])){
                top++;
                cs[top] = cs[i];// 如果只有一种括号就不需要这行
            }
            else
                top--;
        }
        return top == -1;
    }
    
    public boolean ism(char c1, char c2){
        if(c1 == '(' && c2 ==')')
            return true;
        if(c1 == '[' && c2 ==']')
            return true;
        if(c1 == '{' && c2 =='}')
            return true;
        return false;
    }
}