棒球比赛

You are keeping the scores for a baseball game with strange rules. At the beginning of the game, you start with an empty record.

You are given a list of strings operations, where operations[i] is the ith operation you must apply to the record and is one of the following:

• An integer x.
  • Record a new score of x.
• '+'.
  • Record a new score that is the sum of the previous two scores.
• 'D'.
  • Record a new score that is the double of the previous score.
• 'C'.
  • Invalidate the previous score, removing it from the record.

Return the sum of all the scores on the record after applying all the operations.

The test cases are generated such that the answer and all intermediate calculations fit in a 32-bit integer and that all operations are valid.

 

Example 1:

Input: ops = ["5","2","C","D","+"]
Output: 30
Explanation:
"5" - Add 5 to the record, record is now [5].
"2" - Add 2 to the record, record is now [5, 2].
"C" - Invalidate and remove the previous score, record is now [5].
"D" - Add 2 * 5 = 10 to the record, record is now [5, 10].
"+" - Add 5 + 10 = 15 to the record, record is now [5, 10, 15].
The total sum is 5 + 10 + 15 = 30.

Example 2:

Input: ops = ["5","-2","4","C","D","9","+","+"]
Output: 27
Explanation:
"5" - Add 5 to the record, record is now [5].
"-2" - Add -2 to the record, record is now [5, -2].
"4" - Add 4 to the record, record is now [5, -2, 4].
"C" - Invalidate and remove the previous score, record is now [5, -2].
"D" - Add 2 * -2 = -4 to the record, record is now [5, -2, -4].
"9" - Add 9 to the record, record is now [5, -2, -4, 9].
"+" - Add -4 + 9 = 5 to the record, record is now [5, -2, -4, 9, 5].
"+" - Add 9 + 5 = 14 to the record, record is now [5, -2, -4, 9, 5, 14].
The total sum is 5 + -2 + -4 + 9 + 5 + 14 = 27.

Example 3:

Input: ops = ["1","C"]
Output: 0
Explanation:
"1" - Add 1 to the record, record is now [1].
"C" - Invalidate and remove the previous score, record is now [].
Since the record is empty, the total sum is 0.

 

Constraints:

• 1 <= operations.length <= 1000
• operations[i] is "C", "D", "+", or a string representing an integer in the range [-3 * 104, 3 * 104].
• For operation "+", there will always be at least two previous scores on the record.
• For operations "C" and "D", there will always be at least one previous score on the record.

/*
 * 题目思路：
 * 1. 创建一个列表来存储每次得分的记录。
 * 2. 遍历 ops 数组，根据不同的操作类型更新记录：
 *    - 整数：直接加入记录。
 *    - 'C'：移除最后一次得分。
 *    - 'D'：加入最后一次得分的两倍。
 *    - '+'：加入最后两次得分的和。
 * 3. 使用 Java Stream API 计算记录的总和。
 */

import java.util.ArrayList;
import java.util.List;

public class BaseballGameStream {
    public int calPoints(String[] ops) {
        List<Integer> record = new ArrayList<>();
        for (String op : ops) {
            switch (op) {
                case "C":
                    record.remove(record.size() - 1);
                    break;
                case "D":
                    record.add(record.get(record.size() - 1) * 2);
                    break;
                case "+":
                    record.add(record.get(record.size() - 1) + record.get(record.size() - 2));
                    break;
                default:
                    record.add(Integer.parseInt(op));
                    break;
            }
        }
        return record.stream().mapToInt(Integer::intValue).sum();
    }
}

该题解使用一个栈（列表）来模拟整个计分过程。遍历给定的操作列表，根据不同的操作符执行相应的计分操作。对于整数，直接将其加入到总分中，并将其压入栈中；对于 'C' 操作，从总分中减去栈顶元素，并将栈顶元素弹出；对于 'D' 操作，将栈顶元素的两倍加入到总分中，并将其压入栈中；对于 '+' 操作，将栈顶两个元素的和加入到总分中，并将其压入栈中。最后返回总分即可。

O(n)

O(n)