leetcode刷题（20）——判断括号的有效性

最新推荐文章于 2024-11-19 23:58:32 发布

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最新推荐文章于 2024-11-19 23:58:32 发布

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本文介绍了一种使用栈数据结构来验证括号是否正确配对的方法。通过遍历输入字符串，遇到开括号将其压入栈中，遇到闭括号则检查栈顶元素是否为对应的开括号并弹出。最后，若栈为空则说明所有括号均正确配对。

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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

leetcode中不需要导包

class Solution {
    public boolean isValid(String s) {
        Stack<Character> sk = new Stack<>();
        Character c;
        for(int i=0;i<s.length();i++){
            c = s.charAt(i);  //字符串常用函数
            if(c=='('||c=='['||c=='{'){sk.push(c);}
            //注意!sk.isEmpty()是为了防止字符串后面多出括号')'出现无效
            else if(c==')'&&!sk.isEmpty()&&sk.peek()=='('){sk.pop();}  
            else if(c==']'&&!sk.isEmpty()&&sk.peek()=='['){sk.pop();}
            else if(c=='}'&&!sk.isEmpty()&&sk.peek()=='{'){sk.pop();}
            else{return false;}
        }
        return sk.isEmpty();  //栈刚好为空则正好匹配，不为空则证明字符串后面少括号
    }
}