【LeetCode】 122 有效的括號

題目：

解題思路：

stack.peek() 返回棧頂元素，但不在堆棧中刪除它。 Stack.pop() 返回棧頂元素，並在進程中刪除它。

https://leetcode-cn.com/problems/valid-parentheses/solution/you-xiao-de-gua-hao-by-leetcode-solution/

代碼：

import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;

public class LC129 {

    public boolean isValid(String s) {
        int n = s.length();
        if (n % 2 == 1) {
            return false;
        }

        Map<Character, Character> pairs = new HashMap<Character, Character>() {{
            put(')', '(');
            put(']', '[');
            put('}', '{');
        }};
        Deque<Character> stack = new LinkedList<Character>();
        for (int i = 0; i < n; i++) {
            char ch = s.charAt(i);
            if (pairs.containsKey(ch)) {
                if (stack.isEmpty() || stack.peek() != pairs.get(ch)) {
                    return false;
                }
                stack.pop();
            } else {
                stack.push(ch);
            }
        }
        return stack.isEmpty();
    }

    public static void main(String[] args) {
        LC129 lc = new LC129();
        System.out.println(lc.isValid("(){}[]"));
    }
}