495. Teemo Attacking（python+cpp）

题目：

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.

解释：
注意，可以在敌方已经中毒的前提下再次攻击使其再次中毒，也就是说，如果在对方还是中毒状态下就再次使其中毒，对方中毒的总时间会缩短，中毒的效果并不会叠加。
每次加上duration和 timeSeries[i+1]-timeSeries[i]中较小的一个即可。
注意，即使是可以重复中毒，但是duration的时间是不变的，所以下次中毒结束之前，本次中毒的效果一定会结束，所以问题没有那么复杂。
python代码：

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++代码：

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;    
    }
};

总结：