[LeetCode] Baseball Game

Posted 2020-10-09 immjc

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‘s points.
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: 
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.

这是一个棒球计分问题，需要对string中的每一个元素进行判断，如果该元素是数字，则存储在结果数组中，并计算当前的总分数，如果是字母，则是对已有的分数进行操作。

大致思路是：利用一个vector来存储当前的除了操作符的数字，用一个res变量计算当前的总分数。遍历每一个ops元素并判断，然后对vector的元素进行操作即可，需要注意的是要实时更新结果res和记录vector中末尾元素的索引。

使用vector来存储每一条指令的数字结果。利用pop_back()来处理数组最后一个元素。

class Solution {
public:
    int calPoints(vector<string>& ops) {
        vector<int> val;
        int i = 0, res = 0;
        for (auto s : ops) {
            if (s == "+") {
                int tmp1 = val[i - 1] + val[i - 2];
                res += tmp1;
                val.push_back(tmp1);
                i++;
            }
            else if (s == "C") {
                int tmp2 = val.back();
                res -= tmp2;
                val.pop_back();
                i--;
            }
            else if (s == "D") {
                int tmp3 = val[i - 1] * 2;
                res += tmp3;
                val.push_back(tmp3);
                i++;
            }
            else {
                int num = stoi(s);
                res += num;
                val.push_back(num);
                i++;
            }
        }
        return res;
    }
};
// 6 ms