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?

跟Leetcode里search for a range挺像的，就是找到一个target之后，还要继续找它的左边沿。最后l指针超过r指针之后， l 指针会停在左边沿上

 class Solution {
     /**
      * @param nums: The integer array.
      * @param target: Target to find.
      * @return: The first position of target. Position starts from 0.
      */
     public int binarySearch(int[] nums, int target) {
         int l = 0, r = nums.length - 1;
         int m = 0;
         while (l <= r) {
             m = (l + r) / 2;
             if (nums[m] == target) break;
             else if (nums[m] > target) r = m - 1;
             else l = m + 1;
         }
         if (nums[m] != target) return -1;
         l = 0;
         r = m;
         while (l <= r) {
             m = (l + r) / 2;
             if (nums[m] == target) {
                 r = m - 1;
             }
             else l = m + 1;
         }
         return l;
     }
 }