Description
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.
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.
4/23/2017
算法班，直接套用解题格式
注意第16，17行把超越输入范围的也包含进去，否则最后一行需要用绝对值来判断。
1 public int totalOccurance(int[] A, int target) { 2 if (A == null || A.length == 0) return -1; 3 int ret; 4 int start = 0, end = A.length - 1; 5 6 while (start + 1 < end) { 7 int mid = start + (end - start) / 2; 8 if (A[mid] == target) return mid; 9 if (A[mid[ < target) { 10 end = mid; 11 } else { 12 start = mid; 13 } 14 } 15 16 if (A[start] >= target) return start; 17 if (A[end] <= target) return end; 18 return Math.abs(A[end] - target) < Math.abs(target - A[start])? end: start; 19 }