题目 #

You are given an integer array nums and two integers minK and maxK.

A fixed-bound subarray of nums is a subarray that satisfies the following conditions:

• The minimum value in the subarray is equal to minK.
• The maximum value in the subarray is equal to maxK.

Return the number of fixed-bound subarrays.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,3,5,2,7,5], minK = 1, maxK = 5
Output: 2
Explanation: The fixed-bound subarrays are [1,3,5] and [1,3,5,2].

Example 2:

Input: nums = [1,1,1,1], minK = 1, maxK = 1
Output: 10
Explanation: Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.

Constraints:

• $2 <= nums.length <= 10^5$
• $1 <= nums[i], minK, maxK <= 10^6$

思路1 分析性质一次遍历 #

分析 #

• 定界子数组里面只有minK和maxK之间的数，并且要求包含minK和maxK
• 可以找一个起点和终点，如果中间包含minK和maxK就代表一个满足条件的子数组，起点到第一个minK或maxK之间的每个点都可以和终点组成一个子数组
• 如果出现一个大于maxK或小于minK的，就将起点设置成下一个
• minK和maxK所在索引都大于起点，就可以计算子数组数量

代码实现 #

 1func min(a, b int) int {
 2	if a < b {
 3		return a
 4	}
 5	return b
 6}
 7
 8func max(a, b int) int {
 9	if a > b {
10		return a
11	}
12	return b
13}
14
15func countSubarrays(nums []int, minK int, maxK int) int64 {
16	s, minI, maxI := -1, -1, -1
17	var result int64 = 0
18	for e, v := range nums {
19		if v < minK || v > maxK {
20			s = e // 子数组不包含s
21			continue
22		}
23		if v == minK {
24			minI = e
25		}
26		if v == maxK {
27			maxI = e
28		}
29		// 1 2 3 4 5
30		// s   i   e
31		// minI或maxI有一个小于s就不用计算，也就是+0
32		// 都大于s，就是起点到较小的那个之间的数量都满足要求
33		result += int64(max(min(minI, maxI)-s, 0))
34	}
35	return result
36}