First Position of Target(lintcode 14)

Description

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.

Interface

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
    }
}

Solution

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 -1;
        }

        int start = 0;
        int end = nums.length - 1;

        while (start + 1 < end) {
            int mid = start + (end - start) / 2;

            if (nums[mid] >= target) {
                end = mid;
            } else {
                start = mid;
            }
        }

        if (nums[start] == target) {
            return start;
        } else if (nums[end] == target) {
            return end;
        } else {
            return -1;
        }
    }
}