题目描述：

LeetCode 495. Teemo Attacking

In LLP world, there is a hero called Teemo and his attacking can make his enemy Ashe be in poisoned condition. Now, given the Teemo's attacking ascending time series towards Ashe and the poisoning time duration per Teemo's attacking, you need to output the total time that Ashe is in poisoned condition.

You may assume that Teemo attacks at the very beginning of a specific time point, and makes Ashe be in poisoned condition immediately.

Example 1:

Input: [1,4], 2
Output: 4
Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned immediately. 
This poisoned status will last 2 seconds until the end of time point 2. 
And at time point 4, Teemo attacks Ashe again, and causes Ashe to be in poisoned status for another 2 seconds. 
So you finally need to output 4.

Example 2:

Input: [1,2], 2
Output: 3
Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned. 
This poisoned status will last 2 seconds until the end of time point 2. 
However, at the beginning of time point 2, Teemo attacks Ashe again who is already in poisoned status. 
Since the poisoned status won't add up together, though the second poisoning attack will still work at time point 2, it will stop at the end of time point 3. 
So you finally need to output 3.

Note:

1. You may assume the length of given time series array won't exceed 10000.
2. You may assume the numbers in the Teemo's attacking time series and his poisoning time duration per attacking are non-negative integers, which won't exceed 10,000,000.

题目大意：

给定一组递增的时间起点timeSeries，以及一个时间段duration，timeSeries中的每个起点st对应的终点ed = st + duration。

求各时间段覆盖的时间总长度。

注意：

1. 你可以假设给定时间序列的长度不超过10000。
2. 你可以假设时间序列中的元素以及时间段长度均为非负整数，并且不超过10000000。

解题思路：

一趟遍历即可，详见代码。

Python代码：

class Solution(object):
    def findPoisonedDuration(self, timeSeries, duration):
        """
        :type timeSeries: List[int]
        :type duration: int
        :rtype: int
        """
        now = ans = 0
        for st in timeSeries:
            ans += min(duration, st + duration - now)
            now = st + duration
        return ans

 

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