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，然后做一個(gè)函數(shù)，去判斷 i是否在set的±k的區(qū)間內(nèi)，在的話，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.