495. Teemo Attacking

In LOL 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.

求英雄中毒的有效时长。

思路：中毒最长持续的时间为duration，但是有可能出现，上次中毒后的药性，在本次仍然持续一段时间，因此，单次中毒的有效时长t = Math.min(duration, timeSeries[i] + duration - end)，其中end为上次中毒持续到的时刻。依次更新end即可，程序如下所示：

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        if (timeSeries.length == 0){
            return 0;
        }
        int res = duration, end = timeSeries[0] + duration;
        for (int i = 1; i < timeSeries.length; ++ i){
            res += Math.min(duration, timeSeries[i] + duration - end);
            end = Math.max(timeSeries[i] + duration, end);
        }
        return res;
    }
}