Given a?sorted?integer array?arr, two integers?k?and?x, return the?k?closest integers to?x?in the array. The result should also be sorted in ascending order.

An integer?a?is closer to?x?than an integer?b?if:

• |a - x| < |b - x|, or

• |a - x| == |b - x|?and?a < b

?

Example 1:

Input: arr = [1,2,3,4,5], k = 4, x = 3

Output: [1,2,3,4]

Example 2:

Input: arr = [1,2,3,4,5], k = 4, x = -1

Output: [1,2,3,4]

?

Constraints:

• 1 <= k <= arr.length

• 1 <= arr.length <= 104

• arr?is sorted in?ascending?order.

• -104?<= arr[i], x <= 104

我是先把跟x的差值的絕對值放到二維數(shù)組中，然后對應(yīng)的值放到另一個位置上，依次排序，

然后找出最接近的k個數(shù)值即可，然后把數(shù)字放到list中就行，其實多操作了一次，沒必要再放到k長度的數(shù)組中，直接放到list中即可。

下面是代碼：

Runtime:?25 ms, faster than?23.48%?of?Java?online submissions for?Find K Closest Elements.

Memory Usage:?43.7 MB, less than?98.88%?of?Java?online submissions for?Find K Closest Elements.