[Leetcode] 682. Baseball Game 解题报告

棒球得分记录器实现

最新推荐文章于 2023-03-30 21:30:31 发布

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本文介绍了一种基于字符串操作的棒球得分记录器实现方案，通过使用C++中的vector来模拟栈的功能，实现了对得分的有效管理和计算。文章详细解释了如何处理不同类型的输入，并给出了具体的示例。

题目：

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.

思路：

练手题目，哈哈。我们可以用vector来完美地模拟stack，而且比stack好的一点是我们可以同时存取里面的任意一个元素。

代码：

class Solution {
public:
    int calPoints(vector<string>& ops) {
        vector<int> scores;     // we use vector to simulate stack
        int sum = 0;
        for (auto &s : ops) {
            if (s == "C") {
                sum -= scores.back();
                scores.pop_back();
            }
            else if (s == "D") {
                scores.push_back(2 * scores.back());
                sum += scores.back();
            }
            else if (s == "+") {
                int score = scores[scores.size() - 2] + scores.back();
                scores.push_back(score);
                sum += score;
            }
            else {
                int score = stoi(s);
                scores.push_back(score);
                sum += score;
            }
        }
        return sum;
    }
};