lintcode-easy-First Position of Target

For a given sorted array (ascending order) and atarget number, find the first index of this number inO(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, return2.

Challenge

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

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) {
        //write your code here
        if(nums == null || nums.length == 0)
            return 0;
        
        int left = 0;
        int right = nums.length - 1;
        
        while(left < right){
            int mid = left + (right - left) / 2;
            
            if(nums[mid] < target)
                left = mid + 1;
            else if(nums[mid] > target)
                right = mid - 1;
            else
                right = mid;
        }
        
        if(nums[left] == target)
            return left;
        else
            return -1;
    }
}