Add solution 0135

leetcode/0135.Candy/135. Candy.go

@@ -0,0 +1,20 @@
+package leetcode
+
+func candy(ratings []int) int {
+	candies := make([]int, len(ratings))
+	for i := 1; i < len(ratings); i++ {
+		if ratings[i] > ratings[i-1] {
+			candies[i] += candies[i-1] + 1
+		}
+	}
+	for i := len(ratings) - 2; i >= 0; i-- {
+		if ratings[i] > ratings[i+1] && candies[i] <= candies[i+1] {
+			candies[i] = candies[i+1] + 1
+		}
+	}
+	total := 0
+	for _, candy := range candies {
+		total += candy + 1
+	}
+	return total
+}

leetcode/0135.Candy/135. Candy_test.go

@@ -0,0 +1,47 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question135 struct {
+	para135
+	ans135
+}
+
+// para 是参数
+// one 代表第一个参数
+type para135 struct {
+	ratings []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans135 struct {
+	one int
+}
+
+func Test_Problem135(t *testing.T) {
+
+	qs := []question135{
+
+		{
+			para135{[]int{1, 0, 2}},
+			ans135{5},
+		},
+
+		{
+			para135{[]int{1, 2, 2}},
+			ans135{4},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 135------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans135, q.para135
+		fmt.Printf("【input】:%v       【output】:%v\n", p, candy(p.ratings))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0135.Candy/README.md

@@ -0,0 +1,74 @@
+# [135. Candy](https://leetcode.com/problems/candy/)
+
+
+## 题目
+
+There are `n` children standing in a line. Each child is assigned a rating value given in the integer array `ratings`.
+
+You are giving candies to these children subjected to the following requirements:
+
+- Each child must have at least one candy.
+- Children with a higher rating get more candies than their neighbors.
+
+Return *the minimum number of candies you need to have to distribute the candies to the children*.
+
+**Example 1:**
+
+```
+Input: ratings = [1,0,2]
+Output: 5
+Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
+```
+
+**Example 2:**
+
+```
+Input: ratings = [1,2,2]
+Output: 4
+Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
+The third child gets 1 candy because it satisfies the above two conditions.
+```
+
+**Constraints:**
+
+- `n == ratings.length`
+- `1 <= n <= 2 * 10^4`
+- `0 <= ratings[i] <= 2 * 10^4`
+
+## 题目大意
+
+老师想给孩子们分发糖果，有 N 个孩子站成了一条直线，老师会根据每个孩子的表现，预先给他们评分。你需要按照以下要求，帮助老师给这些孩子分发糖果：
+
+- 每个孩子至少分配到 1 个糖果。
+- 评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果。
+
+那么这样下来，老师至少需要准备多少颗糖果呢？
+
+## 解题思路
+
+- 本题的突破口在于，评分更高的孩子必须比他两侧的邻位孩子获得更多的糖果，这句话。这个规则可以理解为 2 条规则，想象成按身高排队，站在下标为 0 的地方往后“看”，评分高即为个子高的，应该比前面个子矮(评分低)的分到糖果多；站在下标为 n - 1 的地方往后“看”，评分高即为个子高的，同样应该比前面个子矮(评分低)的分到糖果多。你可能会有疑问，规则都是一样的，为什么会出现至少需要多少糖果呢？因为可能出现评分一样高的同学。扫描数组两次，处理出每一个学生分别满足左规则或右规则时，最少需要被分得的糖果数量。每个人最终分得的糖果数量即为这两个数量的最大值。两次遍历结束，将所有糖果累加起来即为至少需要准备的糖果数。由于每个人至少分配到 1 个糖果，所以每个人糖果数再加一。
+
+## 代码
+
+```go
+package leetcode
+
+func candy(ratings []int) int {
+	candies := make([]int, len(ratings))
+	for i := 1; i < len(ratings); i++ {
+		if ratings[i] > ratings[i-1] {
+			candies[i] += candies[i-1] + 1
+		}
+	}
+	for i := len(ratings) - 2; i >= 0; i-- {
+		if ratings[i] > ratings[i+1] && candies[i] <= candies[i+1] {
+			candies[i] = candies[i+1] + 1
+		}
+	}
+	total := 0
+	for _, candy := range candies {
+		total += candy + 1
+	}
+	return total
+}
+```