每日一题：括号匹配

leetcode20

package ValidParentheses20;

import java.util.HashMap;
import java.util.Stack;

class Solution {

    private HashMap<Character,Character> map = new HashMap<>();
    public Solution(){
        this.map.put(')','(');
        this.map.put('}','{');
        this.map.put(']','[');

    }
    public boolean isValid(String s) {
        Stack<Character> stack1 = new Stack<>();
        for (int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if (this.map.containsKey(c)){
                char top = stack1.empty() ? '#' : stack1.pop();
                if (top != map.get(c)){
                    return false;
                }
            }else {
                stack1.push(c);
            }
        }
        return stack1.empty();
    }

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

心得体会：

1. 使用一个hashmap存储括号匹配，代替switch语句和if else语句。
2. char top = stack1.empty() ? ‘#’ : stack1.pop(); 用“#”代替判断空栈条件，用stack.pop()代替stack.peek();可以少写一条pop语句。
3. 别忘了写else