LeetCode 682. Baseball Game (Easy)

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:
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.
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.

Example1：

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.

Example2：

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.

题目大意：给一个“计分板”（字符串数组），数字表示积分，“+”表示将上两次的积分的和计入总分，“D”表示将上次的积分的2倍计入总分，“C”表示撤销上次积分，问最后的总积分。

思路：用一个栈，将每一次积分都压栈，遇到“C”就弹出，最后遍历栈算总分即可。

c++代码：

class Solution {

public:
    int calPoints(vector<string>& ops) {
        stack<int> st;
        int sum = 0;
        for (auto &op : ops)
        {
            if (op == "C")
            {
                st.pop();
            }
            else if (op == "+")
            {
                int temp = st.top();
                st.pop();
                int s = temp + st.top();
                st.push(temp);
                st.push(s);
            }
            else if (op == "D")
            {
                st.push(2 * st.top());
            }
            else
            {
                st.push(atoi(op.c_str()));
            }
        }
        while (!st.empty())
        {
            sum += st.top();
            st.pop();
        }
        return sum;
    }
};