LintCode Binary Search

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.

Have you met this question in a real interview? Yes

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 {
public:
    /**
     * @param nums: The integer array.
     * @param target: Target number to find.
     * @return: The first position of target. Position starts from 0.
     */
    int binarySearch(vector<int> &array, int target) {
        // write your code here
        long len = array.size();
        long lo = 0;
        long hi = len;
        while (lo < hi) {
            long mid = (lo + hi) / 2;
            if (array[mid] < target) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }

        return (lo == len || array[lo] != target) ? -1 : lo;
    }
};