Saturday, February 8, 2014

Search for a Range (Java)

Leetcode

Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution: because of O(logn), so we should consider about binary search, so how to apply binary search to find the low bound and high bound? We can make target -0.5 for searching low bound and target+0. 5 for the high bound. Be careful the corner case