LeetCode-Baseball Game

Description：
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.

题意：给定一个字符串数组，包含四种不同的情况；计算返回最后总的得分；

如果字符串是由数字组成的，则总分累加上这个数值；
如果字符串是“C”，那么最后一次累加的值无效，需要从总分中减去这部分；
如果字符串是“D”，那么总分累加上最后一个有效数字的两倍（记为doubleValue），并且这个时候doubleValue成为最后一个有效数值；
如果字符串是“+”，那么总分累加上最后两个有效数字的和（记为plusValue），并且这个时候plusValue成为最后一个有效数值；

解法：要完成上述的操作，我们可以利用栈来模拟这个过程，栈顶保存的总是最后一个有效数字；

Java

class Solution {
    public int calPoints(String[] ops) {
        int result = 0;
        LinkedList<Integer> list = new LinkedList<>();
        for (String op : ops) {
            int value;
            if (Character.isDigit(op.charAt(0)) || op.charAt(0) == '-') {
                value = Integer.valueOf(op);
                list.addLast(value);
                result += value;
            } else if ("C".equals(op)) {
                value = list.removeLast();
                result -= value;
            } else if ("D".equals(op)) {
                value = list.getLast();
                result += 2 * value;
                list.addLast(2 * value);
            } else if ("+".equals(op)) {
                value = list.getLast() + list.get(list.size() - 2);
                result += value;
                list.addLast(value);
            }
        }
        return result;
    }
}

Java

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