LeetCode 2200. Find All K-Distant Indices in an Array

You are given a 0-indexed integer array nums and two integers key and k. A k-distant index is an index i of nums for which there exists at least one index j such that |i - j| <= k and nums[j] == key.

Return a list of all k-distant indices sorted in increasing order.

Example 1:

Input: nums = [3,4,9,1,3,9,5], key = 9, k = 1Output: [1,2,3,4,5,6]Explanation: Here, nums[2] == key and nums[5] == key. - For index 0, |0 - 2| > k and |0 - 5| > k, so there is no j where |0 - j| <= k and nums[j] == key. Thus, 0 is not a k-distant index. - For index 1, |1 - 2| <= k and nums[2] == key, so 1 is a k-distant index. - For index 2, |2 - 2| <= k and nums[2] == key, so 2 is a k-distant index. - For index 3, |3 - 2| <= k and nums[2] == key, so 3 is a k-distant index. - For index 4, |4 - 5| <= k and nums[5] == key, so 4 is a k-distant index. - For index 5, |5 - 5| <= k and nums[5] == key, so 5 is a k-distant index. - For index 6, |6 - 5| <= k and nums[5] == key, so 6 is a k-distant index.Thus, we return [1,2,3,4,5,6] which is sorted in increasing order.

Example 2:

Input: nums = [2,2,2,2,2], key = 2, k = 2Output: [0,1,2,3,4]Explanation: For all indices i in nums, there exists some index j such that |i - j| <= k and nums[j] == key, so every index is a k-distant index. Hence, we return [0,1,2,3,4].

Constraints:

• 1 <= nums.length <= 1000

• 1 <= nums[i] <= 1000

• key is an integer from the array nums.

• 1 <= k <= nums.length

把所有等于Key的index放到set中，然后做一个函数，去判断 i是否在set的±k的区间内，在的话，放到list中，最后返回list；

Runtime: 10 ms, faster than 54.59% of Java online submissions for Find All K-Distant Indices in an Array.

Memory Usage: 44.2 MB, less than 6.99% of Java online submissions for Find All K-Distant Indices in an Array.