题目描述

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.

解法一

思路

如果只有Integer+D的话这题就非常简单了,只用保存上两轮的得分,以及一个累加和就好了。但注意到这里有C可以撤销上一轮的得分,并且C出现的次数没有限制,那么这里就必须用到栈来记录和管理之前每轮的得分了。

那么这题实际上就是对栈的操作了:

  • Integer:本轮得分为Integer,入栈。
  • +:本轮得分为当前两个栈顶值之和,入栈。
  • D:本轮得分为当前栈顶值的两倍,入栈。
  • C:栈顶出栈。

最终结果为当前栈中所有值的和。

Python

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

Java

class Solution {
    public int calPoints(String[] ops) {
        Stack<Integer> history = new Stack<>();
        for (String op : ops) {
            switch (op) {
            case "+":
                history.push(history.peek() + history.get(history.size() - 2));
                break;
            case "D":
                history.push(history.peek() * 2);
                break;
            case "C":
                history.pop();
                break;
            default:
                history.push(Integer.parseInt(op));
            }
        }
        int result = 0;
        for (int item : history) {
            result += item;
        }
        return result;
    }
}

C++

class Solution {
public:
    int calPoints(vector<string> &ops) {
        vector<int> history;
        for (auto &op : ops) {
            if (op == "+") {
                history.push_back(history.back() + history[history.size() - 2]);
            } else if (op == "D") {

                history.push_back(history.back() * 2);
            } else if (op == "C") {
                history.pop_back();
            } else {
                history.push_back(stoi(op));
            }
        }
        return accumulate(history.begin(), history.end(), 0);
    }
};