Add solution 0795

leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum.go

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+package leetcode
+
+func numSubarrayBoundedMax(nums []int, left int, right int) int {
+	return getAnswerPerBound(nums, right) - getAnswerPerBound(nums, left-1)
+}
+
+func getAnswerPerBound(nums []int, bound int) int {
+	res, count := 0, 0
+	for _, num := range nums {
+		if num <= bound {
+			count++
+		} else {
+			count = 0
+		}
+		res += count
+	}
+	return res
+}

leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/795. Number of Subarrays with Bounded Maximum_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question795 struct {
+	para795
+	ans795
+}
+
+// para 是参数
+// one 代表第一个参数
+type para795 struct {
+	nums  []int
+	left  int
+	right int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans795 struct {
+	one int
+}
+
+func Test_Problem795(t *testing.T) {
+
+	qs := []question795{
+
+		{
+			para795{[]int{2, 1, 4, 3}, 2, 3},
+			ans795{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 795------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans795, q.para795
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numSubarrayBoundedMax(p.nums, p.left, p.right))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0795.Number-of-Subarrays-with-Bounded-Maximum/README.md

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+# [795. Number of Subarrays with Bounded Maximum](https://leetcode.com/problems/number-of-subarrays-with-bounded-maximum/)
+
+
+## 题目
+
+We are given an array `nums` of positive integers, and two positive integers `left` and `right` (`left <= right`).
+
+Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least `left` and at most `right`.
+
+```
+Example:Input:
+nums = [2, 1, 4, 3]
+left = 2
+right = 3
+Output: 3
+Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
+```
+
+**Note:**
+
+- `left`, `right`, and `nums[i]` will be an integer in the range `[0, 109]`.
+- The length of `nums` will be in the range of `[1, 50000]`.
+
+## 题目大意
+
+给定一个元素都是正整数的数组`A` ，正整数 `L` 以及 `R` (`L <= R`)。求连续、非空且其中最大元素满足大于等于`L` 小于等于`R`的子数组个数。
+
+## 解题思路
+
+- 题目要求子数组最大元素在 [L,R] 区间内。假设 count(bound) 为计算所有元素都小于等于 bound 的子数组数量。那么本题所求的答案可转化为 count(R) - count(L-1)。
+- 如何统计所有元素小于 bound 的子数组数量呢？使用 count 变量记录在 bound 的左边，小于等于 bound 的连续元素数量。当找到一个这样的元素时，在此位置上结束的有效子数组的数量为 count + 1。当遇到一个元素大于 B 时，则在此位置结束的有效子数组的数量为 0。res 将每轮 count 累加，最终 res 中存的即是满足条件的所有子数组数量。
+
+## 代码
+
+```go
+package leetcode
+
+func numSubarrayBoundedMax(nums []int, left int, right int) int {
+	return getAnswerPerBound(nums, right) - getAnswerPerBound(nums, left-1)
+}
+
+func getAnswerPerBound(nums []int, bound int) int {
+	res, count := 0, 0
+	for _, num := range nums {
+		if num <= bound {
+			count++
+		} else {
+			count = 0
+		}
+		res += count
+	}
+	return res
+}
+```