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:

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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:

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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.

思路

我们可以利用额外的数组空间来储存冷却时间；当然也可以将冷却时间存在一个单独的int变量里面。用额外的数组存储的话相对比较简单，我们只需要记住时间序列之间的差值，并逐一对比即可。如果要用单个变量存储，需要每次都比较差值与duration的关系，即倒数2-5行所做的工作。

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class Solution(object):
    def findPoisonedDuration(self, timeSeries, duration):
        """
        :type timeSeries: List[int]
        :type duration: int
        :rtype: int
        """
        times = []
        length = len(timeSeries)
        if (length):
            for i in range(length-1):
                times.append(timeSeries[i+1] - timeSeries[i])
        else:
            return 0
        ans = 0
        for i in times:
            if (i >= duration):
                ans += duration
            else:
                ans += i
        return ans + duration