我的leetcode做题记录 leetcode 20: Valid Parentheses_valid parentheses: given a string s containing jus-优快云博客

本文链接：https://blog.youkuaiyun.com/HelenZhaoUS/article/details/117029112

本文介绍了一种使用栈数据结构来验证括号序列有效性的方法。通过遍历输入字符串，将左括号压入栈中，并在遇到右括号时检查栈顶元素是否为匹配的左括号。最后确保栈为空以确认括号序列的有效性。

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Leetcode 20. Valid Parentheses

Given a string s 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.

Example 1:

Input: s = "()"
Output: true

Example 2:

Input: s = "()[]{}"
Output: true

Example 3:

Input: s = "(]"
Output: false

Example 4:

Input: s = "([)]"
Output: false

Example 5:

Input: s = "{[]}"
Output: true

Constraints:

1 <= s.length <= 104
s consists of parentheses only '()[]{}'.

Solution（java）:

class Solution {
    public boolean isValid(String s) {
        char[] symbol = {'(' , ')' , '{' , '}' , '[' , ']' };
        int len = s.length();
        if(len%2 != 0 ) return false;
        
        Stack<Character> parentheses= new Stack<Character>();
        
        for(int i = 0 ; i<s.length(); i++){
            if (s.charAt(i) == symbol[0]) parentheses.push(s.charAt(i));
            if (s.charAt(i) == symbol[2]) parentheses.push(s.charAt(i));
            if (s.charAt(i) == symbol[4]) parentheses.push(s.charAt(i));
            
            if (parentheses.empty()) return false;
           else if (s.charAt(i) == symbol[1] ) {
                if (parentheses.peek() !=symbol[0] ) return false;
                else parentheses.pop();
            }
           else if (s.charAt(i) == symbol[3]) {
                if (parentheses.peek() !=symbol[2] ) return false;
                else parentheses.pop();
            }
           else if (s.charAt(i) == symbol[5]) {
                if (parentheses.peek() !=symbol[4] ) return false;
                else parentheses.pop();
            }
        }
        return parentheses.empty();
        
    }
}

使用栈来解决该问题：

栈的特点：后进先出

如遇到左半边括号：放入栈中。

如遇到右半边括号：若此时栈为空（parentheses.empty() ），则return false.

否则，判断栈顶元素(parentheses.peek())与当前括号是否匹配，若匹配，则继续循环。若不匹配，则return false.

最后判断栈是否为空，若栈为空，则该问题为true。

参考资料：

https://leetcode.com/problems/valid-parentheses/

https://www.cnblogs.com/grandyang/p/4424587.html