LeetCode 20. 有效的括号

class Solution {
    public boolean isValid(String s) {
        //首先边界判断,如果是奇数个 直接返回flase
        if(s.length() % 2 == 1){
            return false ;
        }
        //将(){} []存到一个HashMap 中 这样查找用O(1)时间
        Map<Character,Character> paris = new HashMap<Character,Character>();
         paris.put(')','(');
         paris.put(']','[');
         paris.put('}','{');

         //用一个辅助栈来保存
         Deque<Character> stack = new LinkedList<Character>();
         //开始遍历字符串
         for(int i=0;i<s.length();i++){
            char ch = s.charAt(i);
            if(paris.containsKey(ch)){
                if(stack.isEmpty() || stack.peek() != paris.get(ch)){
                    return false ;
                }
                stack.pop();
            }else {
                stack.push(ch);
            }
         }
         return stack.isEmpty();
     }
}
原文地址:https://www.cnblogs.com/peanut-zh/p/13891269.html