lintcode459- Closest Number in Sorted Array- easy

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.

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

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.

用二分法模板写，想想可以知道最后一定在胜出的start,end两数之间，判断。

public class Solution {
    /*
     * @param A: an integer array sorted in ascending order
     * @param target: An integer
     * @return: an integer
     */
    public int closestNumber(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0){
            return -1;
        }

        int start = 0;
        int end = A.length - 1;

        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (target == A[mid]){
                return mid;
            }
            else if (target < A[mid]){
                end = mid;
            }
            else {
                start = mid;
            }
        }

        if (Math.abs(A[start] - target) <= Math.abs(A[end] - target)){
            return start;
        }
        else {
            return end;
        }
    }
}