Merge pull request #187 from gostool/leetcode0495

Leetcode0495

leetcode/0495.Teemo-Attacking/495.Teemo Attacking.go

@@ -0,0 +1,16 @@
+package leetcode
+
+func findPoisonedDuration(timeSeries []int, duration int) int {
+	var ans int
+	for i := 1; i < len(timeSeries); i++ {
+		t := timeSeries[i-1]
+		end := t + duration - 1
+		if end < timeSeries[i] {
+			ans += duration
+		} else {
+			ans += timeSeries[i] - t
+		}
+	}
+	ans += duration
+	return ans
+}

leetcode/0495.Teemo-Attacking/495.Teemo Attacking_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question495 struct {
+	para495
+	ans495
+}
+
+// para 是参数
+type para495 struct {
+	timeSeries []int
+	duration   int
+}
+
+// ans 是答案
+type ans495 struct {
+	ans int
+}
+
+func Test_Problem495(t *testing.T) {
+
+	qs := []question495{
+
+		{
+			para495{[]int{1, 4}, 2},
+			ans495{4},
+		},
+
+		{
+			para495{[]int{1, 2}, 2},
+			ans495{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 495------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans495, q.para495
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findPoisonedDuration(p.timeSeries, p.duration))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0495.Teemo-Attacking/README.md

@@ -0,0 +1,83 @@
+# [495. Teemo Attacking](https://leetcode-cn.com/problems/teemo-attacking/)
+
+
+## 题目
+
+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 <= 10000
+- 0 <= timeSeries[i], duration <= 10000000
+- timeSeries is sorted in non-decreasing order.
+
+## 题目大意
+
+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。
+
+当提莫攻击艾希，艾希的中毒状态正好持续duration 秒。
+
+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。
+
+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。
+
+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 。
+
+返回艾希处于中毒状态的总秒数。
+
+## 解题思路
+
+- i从1开始计数，令t等于timeSeries[i - 1]
+- 比较end(t + duration - 1)和timeSeries[i]的大小，
+  - 如果end小于timeSeries[i],ans+=duration
+  - 否则ans += timeSeries[i] - t
+- ans += duration并返回ans
+
+## 代码
+
+```go
+package leetcode
+
+func findPoisonedDuration(timeSeries []int, duration int) int {
+	var ans int
+	for i := 1; i < len(timeSeries); i++ {
+		t := timeSeries[i-1]
+		end := t + duration - 1
+		if end < timeSeries[i] {
+			ans += duration
+		} else {
+			ans += timeSeries[i] - t
+		}
+	}
+	ans += duration
+	return ans
+}
+```