495. Teemo Attacking

Our hero Teemo is attacking an enemy Ashe with poison attacks! When Teemo attacks Ashe, Ashe gets poisoned for a exactly duration seconds. More formally, an attack at second t will mean Ashe is poisoned during the inclusive time interval [t, t + duration - 1]. If Teemo attacks again before the poison effect ends, the timer for it is reset, and the poison effect will end duration seconds after the new attack.

You are given a non-decreasing integer array timeSeries, where timeSeries[i] denotes that Teemo attacks Ashe at second timeSeries[i], and an integer duration.

Return the total number of seconds that Ashe is poisoned.

Example 1:

Input: timeSeries = [1,4], duration = 2
Output: 4
Explanation: Teemo's attacks on Ashe go as follows:
- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
- At second 4, Teemo attacks, and Ashe is poisoned for seconds 4 and 5.
Ashe is poisoned for seconds 1, 2, 4, and 5, which is 4 seconds in total.

Example 2:

Input: timeSeries = [1,2], duration = 2
Output: 3
Explanation: Teemo's attacks on Ashe go as follows:
- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
- At second 2 however, Teemo attacks again and resets the poison timer. Ashe is poisoned for seconds 2 and 3.
Ashe is poisoned for seconds 1, 2, and 3, which is 3 seconds in total.

Constraints:

• 1 <= timeSeries.length <= 104
• 0 <= timeSeries[i], duration <= 107
• timeSeries is sorted in non-decreasing order.

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这道题好像没啥技巧……就顺着思路写。遍历数组，如果当前数字的下一个比它加上duration大，那就能attack到duration那么长时间，否则就只attack到下一个数字-当前数字的时间。因为最后一次必定能attack到duration那么长时间，所以遍历到倒数第二个就行，最后无脑加上duration返回。

Runtime: 4 ms, faster than 63.73% of Java online submissions for Teemo Attacking.

Memory Usage: 54 MB, less than 65.24% of Java online submissions for Teemo Attacking.

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int result = 0;
        int i = 0;
        while (i < timeSeries.length - 1) {
            int end = timeSeries[i] + duration;
            if (timeSeries[i + 1] > end) {
                result += duration;
            } else {
                result += (timeSeries[i + 1] - timeSeries[i]);
            }
            i++;
        }
        return result + duration;
    }
}

然后看了discussion发现有人提出可以采用merge interval的思想，用两个变量start和end来记录当前attack的开始和结束时间。如果当前数字比end大，说明此次attack结束，result就加上end - start的时间并更新start和end；否则说明此次attack还没结束，只需要更新end。贴一个别人写的代码，就懒的自己写了……Loading...