in LintCode Array Binary Search ~ read.

First Position of Target

LintCode-14.First Position of Target

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 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) {
        if (nums == null || nums.length == 0) {
            return -1;
        }

        int low = 0, high = nums.length - 1;

        while (low + 1 < high) {
            int middle = low + (high - low) / 2;

            if (nums[middle] == target) {
                high = middle;
            } else if (nums[middle] > target) {
                high = middle;
            } else {
                low = middle;
            }
        }

        if (nums[low] == target) {
            return low;
        }
        if (nums[high] == target) {
            return high;
        }

        return -1;
    }
}

Hope this helps,
Michael