*Binary Search

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

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Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

Challenge

If the count of numbers is bigger than 2^32, can your code work properly?

 1     public int binarySearch(int[] nums, int target) 
 2     {
 3         //write your code here
 4         if(nums.length==0)return -1;
 5         int left = 0;
 6         int right = nums.length-1;
 7         while(left+1 <right)
 8         {
 9             int mid = (left + right) /2;
10             if(target==nums[mid])
11             {
12                 right = mid;
13             }
14             
15             if(target<nums[mid])
16             {
17                 right = mid-1;
18             }
19             if(target>nums[mid])
20             {
21                 left = mid+1;
22             }
23             
24         }
25         
26         if(nums[left]==target) return left;
27         if(nums[right]==target) return right;
28         return -1;
29     }