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;
    }
}