LeetCode 682. Baseball Game 棒球比赛(C++/Java)

题目：

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.

分析：

记录棒球比赛的比分，有如下几个规则：

如果是数字的话，该轮等分就是这个数字。

如果是D，该轮得分等于上一轮得分的双倍。

如果是+，该轮的分等于前两轮得分之和。

如果是C，前一轮得分错误，应该删去前一轮得分。

知道所有规则，题目就简单多了，用一个数组来保存每一轮的得分，依据所给的操作，对数组进行相应的操作即可，最后返回数组的和。

程序：

C++

class Solution {
public:
    int calPoints(vector<string>& ops) {
        vector<int> res;
        for(string s:ops){
            int n = res.size();
            if(s == "+"){
                res.push_back(res[n-1] + res[n-2]);
            }else if(s == "D"){
                res.push_back(2 * res[n-1]);
            }else if(s == "C"){
                res.pop_back();
            }else{
                res.push_back(atoi(s.c_str()));
            }
        }
        int sum = 0;
        for(int i:res){
            sum += i;
        }
        return sum;
    }
};

Java

class Solution {
    public int calPoints(String[] ops) {
        List<Integer> list = new ArrayList<>();
        for(String s:ops){
            int n = list.size();
            if(s.equals("+")){
                list.add(list.get(n-1) + list.get(n-2));
            }else if(s.equals("D")){
                list.add(2 * (int)list.get(n-1));
            }else if(s.equals("C")){
                list.remove(n-1);
            }else{
                list.add(Integer.parseInt(s));
            }
        }
        int sum = 0;
        for(int i:list)
            sum += i;
        return sum;
    }
}