題目：

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:
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.

解釋：
注意，可以在敵方已經中毒的前提下再次攻擊使其再次中毒，也就是說，如果在對方還是中毒狀態下就再次使其中毒，對方中毒的總時間會縮短，中毒的效果並不會疊加。
每次加上duration和 timeSeries[i+1]-timeSeries[i]中較小的一個即可。
注意，即使是可以重複中毒，但是duration的時間是不變的，所以下次中毒結束之前，本次中毒的效果一定會結束，所以問題沒有那麼複雜。
python程式碼：

class Solution(object):
    def findPoisonedDuration(self, timeSeries, duration):
        """
        :type timeSeries: List[int]
        :type duration: int
        :rtype: int
        """
        _len=len(timeSeries)
        if _len==0:
            return 0
        if _len ==1:
            return duration
        _sum=0
        for i in range(_len-1):
            _sum+=min(duration,timeSeries[i+1]-timeSeries[i])
        return _sum+duration

c++程式碼：

class Solution {
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
        int n=timeSeries.size();
        if (n==0)
            return 0;
        if (n==1)
            return duration;
        int result=0;
        for (int i=0;i<n-1;i++)
            result+=min(duration,timeSeries[i+1]-timeSeries[i]);
        return result+duration;    
    }
};

總結：