Add to List 658. Find K Closest Elements

Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

Example 1:

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

Example 2:

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

Note:

The value k is positive and will always be smaller than the length of the sorted array.Length of the given array is positive and will not exceed 104Absolute value of elements in the array and x will not exceed 104

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class Solution { public List<Integer> findClosestElements(List<Integer> arr, int k, int x) { if (x <= arr.get(0)) return arr.subList(0, k); if (x >= arr.get(arr.size() - 1)) return arr.subList(arr.size() - k, arr.size()); LinkedList<Integer> re = new LinkedList<>(); int lo = 0; int hi = arr.size() - 1; int i = -1; while (lo <= hi) { int mid = lo + (hi - lo) / 2; if (arr.get(mid) == x) { i = mid; break; } else if (arr.get(mid) < x) lo = mid + 1; else hi = mid - 1; } if (i == -1) { int lo_gap = Math.abs(arr.get(lo) - x); int hi_gap = Math.abs(arr.get(hi) - x); i = lo_gap > hi_gap ? hi : lo; } re.add(arr.get(i)); lo = i - 1; hi = i + 1; while (--k > 0) { int lo_gap = Integer.MAX_VALUE; int hi_gap = Integer.MAX_VALUE; if (lo >= 0) lo_gap = Math.abs(arr.get(lo) - x); if (hi < arr.size()) hi_gap = Math.abs(arr.get(hi) - x); if (lo_gap <= hi_gap) { re.addFirst(arr.get(lo)); lo--; } else { re.addLast(arr.get(hi)); hi++; } } return re; } }