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LeetCode.682.Baseball Game


题目描述

You’re now a baseball game point recorder.

Given a list of strings, each string can be one of the 4 following types:

  1. Integer (one round’s score): Directly represents the number of points you get in this round.
  2. "+" (one round’s score): Represents that the points you get in this round are the sum of the last two valid round’s points.
  3. "D" (one round’s score): Represents that the points you get in this round are the doubled data of the last valid round’s points.
  4. "C" (an operation, which isn’t a round’s score): Represents the last valid round’s points you get were invalid and should be removed.

Each round’s operation is permanent and could have an impact on the round before and the round after.

You need to return the sum of the points you could get in all the rounds.

Example 1:

Input: ["5","2","C","D","+"]
Output: 30
Explanation: 
Round 1: You could get 5 points. The sum is: 5.
Round 2: You could get 2 points. The sum is: 7.
Operation 1: The round 2's data was invalid. The sum is: 5.  
Round 3: You could get 10 points (the round 2's data has been removed). The sum is: 15.
Round 4: You could get 5 + 10 = 15 points. The sum is: 30.

Example 2:

Input: ["5","-2","4","C","D","9","+","+"]
Output: 27
Explanation: 
Round 1: You could get 5 points. The sum is: 5.
Round 2: You could get -2 points. The sum is: 3.
Round 3: You could get 4 points. The sum is: 7.
Operation 1: The round 3's data is invalid. The sum is: 3.  
Round 4: You could get -4 points (the round 3's data has been removed). The sum is: -1.
Round 5: You could get 9 points. The sum is: 8.
Round 6: You could get -4 + 9 = 5 points. The sum is 13.
Round 7: You could get 9 + 5 = 14 points. The sum is 27.

Note:

  • The size of the input list will be between 1 and 1000.
  • Every integer represented in the list will be between -30000 and 30000.

解题过程

开始思考前不小心看到这道题有Stack标签,于是就不用思考了,的确是用栈解决,也很好写,需要注意的地方是"+""D"轮次得到的分数也算是valid分数,不要忘了入栈。

代码如下,刚开始本地运行测试用例没问题,放到LeetCode上Run总是出错,找了半天原因发现犯了个大错,字符串相等判断用了==,对于java中字符串比较要用String.equals()方法还是不习惯啊。但是不知道为啥我本地java1.8的环境用==判断字符串相等竟然没问题。

package _682_BaseballGame;

import java.util.Stack;

class Solution {
    public int calPoints(String[] ops) {
        int sum=0;
        Stack<Integer> stack = new Stack<Integer>();
        for(String op:ops){
            try {
                Integer score = Integer.valueOf(op);
                sum += score;
                stack.push(score);//valid分数入栈
            } catch (NumberFormatException e) {
                if("+".equals(op)){
                    int lastValid = stack.pop();//先将栈顶弹出
                    int lastSecondValid = stack.peek();//取栈顶倒数第二个
                    sum += lastValid + lastSecondValid;
                    stack.push(lastValid);//再将栈顶push进去
                    stack.push(lastValid+lastSecondValid);//valid分数入栈
                }else if("D".equals(op)){
                    sum += stack.peek() * 2;
                    stack.push(stack.peek()*2);//valid分数入栈
                }else if("C".equals(op)){
                    sum -= stack.pop();//弹出栈顶,同时扣分
                }
            }
            System.out.println("sum: "+sum);
        }
        return sum;
    }
}

public class BaseballGame {
    public static void main(String[] args){
        Solution solution = new Solution();
        String[] input1 = {"5","2","C","D","+"};
        String[] input2 = {"5","-2","4","C","D","9","+","+"};
        System.out.println(solution.calPoints(input1));
    }
}

我的代码速度很慢,只打败了2%的人,想着是不是用了try catch导致性能下降,别人的代码好像都没有这么写的,都是用if或switch。用java栈的话代码的确通俗易懂,但性能不行,速度快的代码都是完全手动操作数组。


GitHub代码


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