题目描述：

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:

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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:

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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.

提莫队长，正在送命！

题目比较简单，就是debuff状态时间不会叠加，每次击中目标后重新开始计时，要求返回总的debuff时间。只要每次判断与上一次攻击后的生效时间是否有重合，有的话就减去重合部分即可。

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