495. Teemo Attacking

QUESTIN:

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], 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:

  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.

EXPLANATION:

可以计算出两个攻击之间的间隔,如果超过了duration,那么就加上duration,如果没有超过,就说明被覆盖了,就加上间隔就行,不要忘记的是,最后需要加上最后一击的duration。

SOLUTION:

public class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        if(timeSeries.length==0) return 0;
        int result = 0;
        for(int i = 1;i<timeSeries.length;i++){
            if(timeSeries[i]-timeSeries[i-1]>=duration)
                result+=duration;
            else
                result+=timeSeries[i]-timeSeries[i-1];
        }
        return result+duration;//加上最后一击的持续时间
    }
}
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