# Leetcode 495. Teemo Attacking

Runtime: 1 ms
faster than 100.00%

Memory Usage: 41.6 MB
less than 48.71%

/*
    [2,3,6,8], duration=3
    2 3 4 5 6 7 8 9 10
    3-2=1  => duration 3-1=2
    8-6=2  => duration 3-2=1
    => 2+1=3 少三個(4個數*3(duration)=12 12-3=9) */
    
class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int result=0,minus=0;
        int temp=-1000000000;     //  代表陣列中，前一個數
        
        result = timeSeries.length * duration;
        for(int num : timeSeries){
            if((num-temp)<duration ){
                minus+=duration-(num-temp);
            }
            temp=num;
        }
        
        
        return result-minus;
    }
}

解題思維:
先用全部幾個數去乘duration，再扣掉重複計算的部分
Time: O(n)

###### tags: `Array`