495. Teemo Attacking

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], 2Output: 4Explanation: 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], 2Output: 3Explanation: 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.

Teemo攻击，主要是计算Ashe中毒的总时长，当Ashe在某个时刻中毒后，中毒这种状态会延续duration这么长的时间，在这期间如果再发起攻击,中毒时间就只加1，转换为数组形式的计算就是:当两个相邻元素间的差小于duration时，count就加1，如果差大于等于duration，count加duration，另外到最后一次发起攻击时，中毒状态会延续duration时间，也就是count总是会再加duration，按照这种思想，code如下（timeSeries==null||timeSeries.length==0是为了进行异常处理）：

           int count = 0;       int len = timeSeries.length;              if(timeSeries == null || timeSeries.length == 0){               return 0;           }                     for(int i = 1; i < len; i++){               if(timeSeries[i] - timeSeries[i - 1] >= duration){                   count += duration;               }else{                   count += timeSeries[i] - timeSeries[i - 1];               }           }           count += duration;           return count;