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86 lines
2.9 KiB
Markdown
86 lines
2.9 KiB
Markdown
# [495. Teemo Attacking](https://leetcode.com/problems/teemo-attacking/)
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## 题目
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Our hero Teemo is attacking an enemy Ashe with poison attacks! When Teemo attacks Ashe, Ashe gets poisoned for a exactly duration seconds.
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More formally, an attack at second t will mean Ashe is poisoned during the inclusive time interval [t, t + duration - 1].
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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.
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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.
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Return the total number of seconds that Ashe is poisoned.
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**Example 1**:
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```
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Input: timeSeries = [1,4], duration = 2
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Output: 4
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Explanation: Teemo's attacks on Ashe go as follows:
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- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
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- At second 4, Teemo attacks, and Ashe is poisoned for seconds 4 and 5.
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Ashe is poisoned for seconds 1, 2, 4, and 5, which is 4 seconds in total.
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```
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**Example 2**:
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```
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Input: timeSeries = [1,2], duration = 2
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Output: 3
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Explanation: Teemo's attacks on Ashe go as follows:
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- At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
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- At second 2 however, Teemo attacks again and resets the poison timer. Ashe is poisoned for seconds 2 and 3.
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Ashe is poisoned for seconds 1, 2, and 3, which is 3 seconds in total.
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```
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**Constraints**:
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- 1 <= timeSeries.length <= 10000
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- 0 <= timeSeries[i], duration <= 10000000
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- timeSeries is sorted in non-decreasing order.
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## 题目大意
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在《英雄联盟》的世界中，有一个叫 “提莫” 的英雄。他的攻击可以让敌方英雄艾希（编者注：寒冰射手）进入中毒状态。
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当提莫攻击艾希，艾希的中毒状态正好持续duration 秒。
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正式地讲，提莫在t发起发起攻击意味着艾希在时间区间 [t, t + duration - 1]（含 t 和 t + duration - 1）处于中毒状态。
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如果提莫在中毒影响结束前再次攻击，中毒状态计时器将会重置，在新的攻击之后，中毒影响将会在duration秒后结束。
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给你一个非递减的整数数组timeSeries，其中timeSeries[i]表示提莫在timeSeries[i]秒时对艾希发起攻击，以及一个表示中毒持续时间的整数duration 。
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返回艾希处于中毒状态的总秒数。
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## 解题思路
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- i 从 1 开始计数，令 t 等于 timeSeries[i - 1]
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- 比较 end(t + duration - 1) 和 timeSeries[i] 的大小，
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- 如果 end 小于 timeSeries[i],ans+=duration
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- 否则 ans += timeSeries[i] - t
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- ans += duration 并返回 ans
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## 代码
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```go
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package leetcode
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func findPoisonedDuration(timeSeries []int, duration int) int {
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var ans int
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for i := 1; i < len(timeSeries); i++ {
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t := timeSeries[i-1]
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end := t + duration - 1
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if end < timeSeries[i] {
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ans += duration
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} else {
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ans += timeSeries[i] - t
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}
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}
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ans += duration
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return ans
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}
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```
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