682. Baseball Game-优快云博客

本文介绍了一个棒球得分计算器的设计思路及实现代码。该计算器能够处理四种类型的指令：整数（直接得分）、+（当前得分等于前两次得分之和）、D（当前得分为上次得分的两倍）、C（撤销上次得分）。通过栈结构存储每轮得分，最终计算出所有有效得分的总和。

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You're now a baseball game point recorder.

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

Integer (one round's score): Directly represents the number of points you get in this round.
"+" (one round's score): Represents that the points you get in this round are the sum of the last two valid round's points.
"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.
"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.

题目的大致意思是给你一个字符串

数字表示分数
'+'表示当前轮次的分数等于上两轮分数之和
'D'表示当前轮次的分数等于上一轮分数加倍
'C'表示清除上一次的分数

求最后的总分

思路如下，把字符串转化成分数压入栈中，如果是+就去栈顶的两个元素和然后压入栈中，如果是D，则把栈顶的值乘2再压入栈中

如果是C则把栈顶的元素去除

public int calPoints(String[] ops) {
        Stack<Integer> stack = new Stack();
        for(String s:ops)
        {
            if(s.equals("+"))
            {
                int top = stack.pop();
                int newtop = top+stack.peek();
                stack.push(top);
                stack.push(newtop);
            }
            else if(s.equals("C"))
            {
                stack.pop();
            }
            else if(s.equals("D"))
            {
                 int top = stack.peek();
                 stack.push(2*top);
            }
            else
            {
                 stack.push(Integer.valueOf(s));
            }
        }
        int result = 0;
        while(!stack.empty())
        {
            result = result+stack.pop();
        }
        return result;
    }