Description

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:

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

UPDATE (2017/9/19):
The arr parameter had been changed to an array of integers (instead of a list of integers). Please reload the code definition to get the latest changes.

Solution

Binary search, time O(log n)

I binary-search for where the resulting elements start in the array. It’s the first index i so that arr[i] is better than arr[i+k] (with “better” meaning closer to or equally close to x). Then I just return the k elements starting there.

这思路真心牛逼。。

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int left = 0;
        int right = arr.length - k - 1;
        
        while (left <= right) {
            int mid = (left + right) / 2;
            
            if (x - arr[mid] > arr[mid + k] - x) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        int[] closest = Arrays.copyOfRange(arr, left, left + k);
        return IntStream.of(closest).boxed().collect(Collectors.toList());        
    }
}