Lintcode: First Position of Target (Binary Search)

Binary search is a famous question in algorithm.

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.

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 MAXINT, can your code work properly?

 1 class Solution {
 2     /**
 3      * @param nums: The integer array.
 4      * @param target: Target to find.
 5      * @return: The first position of target. Position starts from 0.
 6      */
 7     public int binarySearch(int[] nums, int target) {
 8         int l = 0, r = nums.length - 1;
 9         int m = 0;
10         while (l <= r) {
11             m = (l + r) / 2;
12             if (nums[m] == target) break;
13             else if (nums[m] > target) r = m - 1;
14             else l = m + 1;
15         }
16         if (nums[m] != target) return -1;
17         l = 0;
18         r = m;
19         while (l <= r) {
20             m = (l + r) / 2;
21             if (nums[m] == target) {
22                 r = m - 1;
23             }
24             else l = m + 1;
25         }
26         return l;
27     }
28 }