Lintcode 460. Find K Closest Elements

Given target, a non-negative integer k and an integer array A sorted in ascending order, find the k closest numbers to target in A, sorted in ascending order by the difference between the number and target. Otherwise, sorted in ascending order by number if the difference is same.

Example

Example 1:

Input: A = [1, 2, 3], target = 2, k = 3
Output: [2, 1, 3]

Example 2:

Input: A = [1, 4, 6, 8], target = 3, k = 3
Output: [4, 1, 6]

Challenge

O(logn + k) time

Notice

1. The value k is a non-negative integer and will always be smaller than the length of the sorted array.
2. Length of the given array is positive and will not exceed $10^4$
3. Absolute value of elements in the array will not exceed $10^4$

Code (Java):

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public class Solution {     /**      * @param A: an integer array      * @param target: An integer      * @param k: An integer      * @return: an integer array      */     public int[] kClosestNumbers(int[] A, int target, int k) {         // write your code here         int[] ans = new int[k];         if (A == null || A.length == 0 || k == 0) {             return ans;         }                   // step 1: find the first number >= target         //         int right = binarySearchGreater(A, target);         int left = right - 1;                   for (int i = 0; i < k; i++) {             if (left < 0) {                 ans[i] = A[right++];             } else if (right >= A.length) {                 ans[i] = A[left--];             } else if (Math.abs(A[left] - target) <= Math.abs(A[right] - target)) {                 ans[i] = A[left--];             } else {                 ans[i] = A[right++];             }         }                   return ans;     }           private int binarySearchGreater(int[] A, int target) {         int start = 0;         int end = A.length - 1;                   while (start + 1 < end) {             int mid = start + (end - start) / 2;             if (A[mid] == target) {                 return mid;             } else if (A[mid] > target) {                 end = mid;             } else {                 start = mid + 1;             }         }                   if (A[start] >= target) {             return start;         }                   if (A[end] >= target) {             return end;         }                   return A.length - 1;     } }