LeetCode-Teemo Attacking

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Description：
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.

题意：在LOL中有一位英雄叫Teemo，他的攻击会使敌人处于中毒的状态，现在给定一个数组表示eemo的攻击时间，每一次攻击敌人的中毒状态会持续duration时间，计算最终敌人中毒的总的时间；

解法：每一次的攻击都会有一个持续的时间duration，并且下一次攻击的时间并不会和上次的时间进行叠加，也就是说，经过下一次的攻击，中毒的持续时间还是duration；因此，我们只需要计算经过 $i$次攻击后，加上持续的时间是否超过 $i+1$的时间点；如果没有超过，那么持续时间只能是duration；否则，持续时间便是 $i+1$次的时间点减去 $i$次的时间点；之后，重复这个过程；

Java

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int res = 0;
        for (int i = 0; i < timeSeries.length - 1; i++) {
            res += timeSeries[i] + duration <= timeSeries[i + 1] ?
                duration : timeSeries[i + 1] - timeSeries[i];
        }
        
        return timeSeries.length == 0 ? 0 : res + duration;
    }
}