[LintCode] Closest Number in Sorted Array

Given a target number and an integer array A sorted in ascending order, find the index i in A such that A[i] is closest to the given target.

Return -1 if there is no element in the array.

Notice

There can be duplicate elements in the array, and we can return any of the indices with same value.

Have you met this question in a real interview?

Yes

Example

Given [1, 2, 3] and target = 2, return 1.

Given [1, 4, 6] and target = 3, return 1.

Given [1, 4, 6] and target = 5, return 1 or 2.

Given [1, 3, 3, 4] and target = 2, return 0 or 1 or 2.

Challenge

O(logn) time complexity.

 1 public class Solution {
 2     /*
 3      * @param A: an integer array sorted in ascending order
 4      * @param target: An integer
 5      * @return: an integer
 6      */
 7     public int closestNumber(int[] A, int target) {
 8         if(A == null || A.length == 0) {
 9             return -1;
10         }
11         int low = 0, high = A.length - 1;
12         while(low + 1 < high) {
13             int mid = low + (high - low) / 2;
14             if(A[mid] == target) {
15                 return mid;
16             }
17             //numbers after index mid are >= A[mid] > target, which means they can't be closer to
18             //the target value than A[mid] is, so we can safely get rid of the bigger half.
19             else if(A[mid] > target) {
20                 high = mid;
21             }
22             //numbers before index mid are <= A[mid] < target, which means they can't be closer to
23             //the target value than A[mid] is, so we can safely get rid of the smaller half.
24             else {
25                 low = mid;
26             }
27         }
28         if(Math.abs(A[low] - target) <= Math.abs(A[high] - target)) {
29             return low;
30         }
31         return high;
32     }
33 }