Merge pull request #214 from gostool/leetcode1518

Leetcode1518

leetcode/1518.Water-Bottles/1518.Water Bottles.go

@@ -0,0 +1,15 @@
+package leetcode
+
+func numWaterBottles(numBottles int, numExchange int) int {
+	if numBottles < numExchange {
+		return numBottles
+	}
+	quotient := numBottles / numExchange
+	reminder := numBottles % numExchange
+	ans := numBottles + quotient
+	for quotient+reminder >= numExchange {
+		quotient, reminder = (quotient+reminder)/numExchange, (quotient+reminder)%numExchange
+		ans += quotient
+	}
+	return ans
+}

leetcode/1518.Water-Bottles/1518.Water Bottles_test.go

@@ -0,0 +1,56 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1518 struct {
+	para1518
+	ans1518
+}
+
+// para 是参数
+type para1518 struct {
+	numBottles  int
+	numExchange int
+}
+
+// ans 是答案
+type ans1518 struct {
+	ans int
+}
+
+func Test_Problem1518(t *testing.T) {
+
+	qs := []question1518{
+
+		{
+			para1518{9, 3},
+			ans1518{13},
+		},
+
+		{
+			para1518{15, 4},
+			ans1518{19},
+		},
+
+		{
+			para1518{5, 5},
+			ans1518{6},
+		},
+
+		{
+			para1518{2, 3},
+			ans1518{2},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1518------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1518, q.para1518
+		fmt.Printf("【input】:%v    【output】:%v\n", p, numWaterBottles(p.numBottles, p.numExchange))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1518.Water-Bottles/README.md

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+# [1518. Water Bottles](https://leetcode-cn.com/problems/water-bottles/)
+
+## 题目
+
+Given numBottles full water bottles, you can exchange numExchange empty water bottles for one full water bottle.
+
+The operation of drinking a full water bottle turns it into an empty bottle.
+
+Return the maximum number of water bottles you can drink.
+
+**Example 1**:
+
+![https://assets.leetcode.com/uploads/2020/07/01/sample_1_1875.png](https://assets.leetcode.com/uploads/2020/07/01/sample_1_1875.png)
+
+    Input: numBottles = 9, numExchange = 3
+    Output: 13
+    Explanation: You can exchange 3 empty bottles to get 1 full water bottle.
+    Number of water bottles you can drink: 9 + 3 + 1 = 13.
+
+**Example 2**:
+
+![https://assets.leetcode.com/uploads/2020/07/01/sample_2_1875.png](https://assets.leetcode.com/uploads/2020/07/01/sample_2_1875.png)
+
+    Input: numBottles = 15, numExchange = 4
+    Output: 19
+    Explanation: You can exchange 4 empty bottles to get 1 full water bottle.
+    Number of water bottles you can drink: 15 + 3 + 1 = 19.
+
+**Example 3**:
+
+    Input: numBottles = 5, numExchange = 5
+    Output: 6
+
+**Example 4**:
+
+    Input: numBottles = 2, numExchange = 3
+    Output: 2
+
+**Constraints:**
+
+- 1 <= numBottles <= 100
+- 2 <= numExchange <= 100
+
+## 题目大意
+
+小区便利店正在促销，用 numExchange 个空酒瓶可以兑换一瓶新酒。你购入了 numBottles 瓶酒。
+
+如果喝掉了酒瓶中的酒，那么酒瓶就会变成空的。
+
+请你计算 最多 能喝到多少瓶酒。
+
+## 解题思路
+
+- 模拟
+首先我们一定可以喝到 numBottles 瓶酒，剩下 numBottles 个空瓶。接下来我们可以拿空瓶子换酒，每次拿出 numExchange 个瓶子换一瓶酒，然后再喝完这瓶酒，得到一个空瓶。这样模拟下去，直到所有的空瓶子小于numExchange结束
+
+## 代码
+
+```go
+package leetcode
+
+func numWaterBottles(numBottles int, numExchange int) int {
+	if numBottles < numExchange {
+		return numBottles
+	}
+	quotient := numBottles / numExchange
+	reminder := numBottles % numExchange
+	ans := numBottles + quotient
+	for quotient+reminder >= numExchange {
+		quotient, reminder = (quotient+reminder)/numExchange, (quotient+reminder)%numExchange
+		ans += quotient
+	}
+	return ans
+}
+```