【两次过】Lintcode 983: Baseball Game

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本文介绍了一种算法，用于解析一系列棒球比赛得分指令，并计算最终得分。该算法使用动态数组存储每轮得分，根据指令类型（直接得分、加倍、累加及取消前次得分）更新得分，并返回所有有效得分的总和。

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.

Notice

The size of the input list will be between 1 and 1000.
Every integer represented in the list will be between -30000 and 30000.

Example

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.

解题思路：

主要是需要建立一个动态数组vector<int>来按规则存放每一轮的分数，最后对vector<int>数组求和即是总分数。

提交第二次才过的原因是，不知道数字字符串应该用什么条件限制，最后才发现只需要将这种情况放在最后的else后即可。

class Solution {
public:
    /**
     * @param ops: the list of operations
     * @return:  the sum of the points you could get in all the rounds
     */
    int calPoints(vector<string> &ops) 
    {
        // Write your code here
        vector<int> result;

        for(int i=0;i<ops.size();i++)
        {
            if(ops[i] == "+")
                result.push_back(*(result.end()-1)+ *(result.end()-2));
            else if(ops[i] == "D")
                result.push_back(*(result.end()-1) + *(result.end()-1));
            else if(ops[i] == "C")
                result.erase(result.end()-1);
            else
                result.push_back(stoi(ops[i]));
        }
        
        int sum=0;
        
        for(auto pt = result.begin();pt != result.end();pt++)
            sum += *pt;

        return sum;
    }
};