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; } } }
lintcode459- Closest Number in Sorted Array- easy
原文：http://www.cnblogs.com/jasminemzy/p/7572675.html