【LeetCode】495. Teemo Attacking【M】【41】

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原创最新推荐文章于 2026-09-07 15:40:11 发布 677 阅读 AI 写同款· GEO 优化 ›

LeetCode题目#495（数组类）——提莫攻击原文链接及题目： https://leetcode-cn.com/problems/max-consecutive-ones/ 我的解法如下： /* 思路：名词转换：攻击事件序列：相当于开始时间；中毒持续时间：持续时间。输入格式为：[第一次攻击起始时间，下一次攻击起始时间]，每次攻击持续时间。需要考虑的点：上一次毒没结束，下一次就开始，那就会产生重叠。所以需要使用不考虑重叠影响情况下算出的数据，减去重叠部分的长度（即前后两个数据求差）也即，需要考虑前一个与后一个元素之间的距离，是否小于攻击持续时间阅读详情

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
        """

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