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.

题目很简单，只需要对每个字符进行判断就行，如果是数字的字符就转换成整型直接相加并存入有效数据里面,C就清除最后一轮有效数据，D就表示当前得分是最后一轮有效数据得分的双倍,+表示你在这一轮得到的得分是最后两个有效回合得分的总和。

 1 int calPoints(char** ops, int opsSize)
 2 {
 3     int sum = 0;
 4     int num = 0;
 5     int k = 0;
 6     int* valid = (int*)malloc(opsSize * sizeof(int)) ;//定义指针，并分配opsSize个sizeof(int)内存单元
 7     memset(valid,0,sizeof(int)*opsSize);
 8     for(int i=0;i<opsSize;i++) {
 9         switch(*ops[i]){
10             case 'C':
11                 sum -= valid[k-1];
12                 valid[--k] = 0;
13                 break;
14             case 'D':
15                 num = valid[k-1] * 2;
16                 valid[k++] = num ;  
17                 sum += num;
18                 break;
19             case '+':
20                 num = valid[k-2] + valid[k-1];
21                 valid[k++] = num ;
22                 sum += num;
23                 break;
24             default:
25                 num = atoi(ops[i]);
26                 sum += num;
27                 valid[k++] = num ; 
28                 break;    
29         }
30 
31     }
32     return sum ;
33 }