[leetcode] 682. Baseball Game @ python

最新推荐文章于 2020-05-28 17:58:59 发布

原创最新推荐文章于 2020-05-28 17:58:59 发布 · 161 阅读

CC 4.0 BY-SA版权

本文详细解析了一种用于棒球比赛计分的算法，该算法能够处理四种类型的输入：整数得分、加号（上两轮得分之和）、大写字母D（上一轮得分的两倍）以及大写字母C（移除上一轮得分）。通过遍历操作列表并根据类型进行相应处理，最终返回所有有效得分的总和。适用于处理大量数据，时间复杂度为O(n)，空间复杂度也为O(n)。

原题

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.

解法

构造列表p储存有效的数字, 遍历ops, 当op为数字时, 将op转化为整数放入p, 其余的按照题意来处理p, 这里注意当分数为负时, op.isdigit()为False, 此时我们要将分数为负的情况加进来.
Time: O(n)
Space: O(n)

代码

class Solution(object):
    def calPoints(self, ops):
        """
        :type ops: List[str]
        :rtype: int
        """
        p = []
        for op in ops:
            if op.isdigit() or (op[0] == '-' and len(op) > 1):
                p.append(int(op))
            elif op == '+':
                p.append(p[-1] + p[-2])
            elif op == 'D':
                p.append(2*p[-1])
            elif op == 'C':
                p.pop()
        return sum(p)