682 Baseball Game

682. Baseball Game

1. Question

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

   validround's points.

3. "D"(one round's score): Represents that the points you get in this round are the doubled data of the last

   validround's points.

4. "C"(an operation, which isn't a round's score): Represents the last

   validround's points you get were invalid and should be removed.

5. 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.

2. Implementation

(1) Stack

class Solution {
    public int calPoints(String[] ops) {
        int sum = 0;
        Stack<Integer> stack = new Stack<>();

        for (String op : ops) {
            if (op.equals("+")) {
                int points1 = stack.pop();
                int points2 = stack.pop();
                int points = points1 + points2;
                sum += points;
                stack.push(points2);
                stack.push(points1);
                stack.push(points);
            }
            else if (op.equals("D")) {
                int points = stack.pop();
                sum += 2 * points;
                stack.push(points);
                stack.push(2 * points);
            }
            else if (op.equals("C")) {
                int points = stack.pop();
                sum -= points;
            }
            else {
                int val = Integer.parseInt(op);
                sum += val;
                stack.push(val);
            }
        }
        return sum;
    }
}

3. Time & Space Complexity

Stack: 时间复杂度O(n), 空间复杂度O(n)