【LeetCode】495. Teemo Attacking【M】【41】

最新推荐文章于 2019-03-25 11:49:53 发布

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本文介绍了一种算法，用于计算游戏角色Teemo攻击对手Ashe导致其处于中毒状态的总时间。通过分析Teemo的攻击时间序列及中毒持续时间，文章提供了具体的实现代码，并通过两个示例解释了算法的工作原理。

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:

You may assume the length of given time series array won't exceed 10000.
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.

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其实就是对每个时间间隔进行判断，要是相邻时间间隔大于duation，那就加上duration，要是小于，那就加上这段时间间隔

后来试了一下，改成min（）来计算了

class Solution(object):
    def findPoisonedDuration(self, timeSeries, duration):

        t = timeSeries
        d = duration
        if not t:
            return 0
        res = 0

        i = 0
        for i in xrange(len(t)-1):

            res += min(d,t[i+1]-t[i])
            '''
            if t[i] + d <= t[i+1]:
                res += d
            else:
                res += t[i+1] - t[i]
            '''
            #i += 1
        res += d
        return res

        """
        :type timeSeries: List[int]
        :type duration: int
        :rtype: int
        """