You’re now a baseball game point recorder. Given a list of strings, each string can be one of the 4 following types: 1.Integer (one round’s score): Directly represents the number of points you get in this round. 2."+" (one round’s score): Represents that the points you get in this round are the sum of the last two valid round’spoints. 3. “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. 4.“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:  
Round1: 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.
 

解釋: 題目要求一目瞭然，用棧做。 python程式碼：

class Solution(object):
    def calPoints(self, ops):
        """
        :type ops: List[str]
        :rtype: int
        """
        lastvalid=0
        valid_list=[]
        for s in ops:
            if s=="C":
                lastvalid=valid_list.pop(-1)
                
            elif s=="D":
                lastvalid=valid_list[-1]
                valid_list.append(2*lastvalid)
            elif s=="+":
                valid_list.append(valid_list[-1]+valid_list[-2])
            else:
                valid_list.append(int(s))
        return sum(valid_list)

c++程式碼：

#include <string>
#include <numeric>
class Solution {
public:
    int calPoints(vector<string>& ops) {
        vector<int> valid_list;
        for (auto op :ops)
        {
            if(op=="C")
                valid_list.pop_back();
            else if(op=="D")
                valid_list.push_back(valid_list[valid_list.size()-1]*2);
            else if(op=="+")
            {  
                valid_list.push_back(valid_list[valid_list.size()-1]+valid_list[valid_list.size()-2]);
            }
            else
                valid_list.push_back(stoi(op));
        }
        return accumulate(valid_list.begin() , valid_list.end() , 0); 
    }
};

總結： 維護一個valid_list ,最後求和即可，不需要每次都更改sum。 accumulate()定義在#include<numeric>中，可以用於累加求和。