LeetCode-682. 棒球比赛

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本文介绍了一种基于栈的棒球游戏计分算法实现，通过解析操作指令，如整数分数、'+'、'D'和'C'，来计算所有有效轮次的总分。算法使用了std::stoi函数和vector容器，详细解释了每种操作的处理方式。

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682. Baseball Game

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.

栈的思想，考虑到如果碰到C，那前一轮得分无效，此时要剔除，如果碰到D，那本轮得分将是前一轮有效得分的两倍，如果碰到‘+’，那本轮得分将是前两轮有效得分的总和，所以可以用栈来保存本轮得分。

stoi()函数可将string转换成int
容器vector做为栈来保存得分

static const int _ = []() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 0;
}();

class Solution {
public:
    int calPoints(vector<string>& ops) {
        int sum = 0;
        vector<int> vec;
        for(int i=0;i<ops.size();i++)
        {
            if(!ops[i].compare("C"))
            {
                int a = vec.back();
                vec.pop_back();
                sum -= a; //减掉前一轮得分
            }
            else if(!ops[i].compare("D"))
            {
                int a = vec.back();
                sum += 2*a;
                vec.push_back(2*a);
            }
            else if(!ops[i].compare("+"))
            {
                //a,b保存前两轮得分
                int a = *(vec.end()-1); 
                int b = *(vec.end()-2);
                sum+=(a+b);
                vec.push_back(a+b);
                
            }
            else
            {
                sum+=stoi(ops[i]);
                vec.push_back(stoi(ops[i]));
            }
        }
        return sum;
    }
};