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=3Output: [1,2,3,4]

Example 2:

Input: [1,2,3,4,5], k=4, x=-1Output: [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 104
Absolute value of elements in the array and x will not exceed 104

code

时间复杂度O(logn)，空间复杂度 O(1)

public class Solution {    public IList<int> FindClosestElements(IList<int> arr, int k, int x) {            IList<int> rtn = new List<int>();            int arrcnt = arr.Count; //count of arr            if (arrcnt == 0) return rtn;            int lo = 0, hi = arrcnt - 1;            //find index with a nearest element to x            while (hi - lo > 1)            {                int mid = lo + (hi - lo)/2;                if (arr[mid] < x)                    lo = mid;                else if (arr[mid] > x)                    hi = mid;                else lo = hi = mid;            }            int nearestind = Math.Abs(arr[lo] - x) <= Math.Abs(arr[hi] - x) ? lo : hi;            //find index range with k nearest elements to x            //surrounding nearestind to find these k             //by backward or forward search            lo = nearestind - 1; hi = nearestind + 1;            int rtncnt = 1; //count of nearest elements to x            while (rtncnt++ < k){                if (hi>=arrcnt)                    lo--;                else if (lo < 0)                    hi++;                else{                    if (Math.Abs(arr[lo] - x) <= Math.Abs(arr[hi] - x))                        lo--;                    else hi++;                }            }            for (int i = lo+1; i < hi; i++)                rtn.Add(arr[i]);            return rtn;    }}