lintcode:First Position of Target

•  Description
•  Notes
•  Testcase
•  Judge

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?

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Related Problems 

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                if (array.size() == 0)            return -1;                    int left = 0;        int right = array.size()-1;                while (left <= right)        {            int mid = left+(right-left)/2;                        if (array[mid] == target)            {                if (mid > 0 && (array[mid-1] != target))                {                    return mid;                }                else if (mid == 0)                {                    return 0;                }                else                {                    right = mid-1;                }            }            else if (array[mid] > target)            {                right = mid-1;            }            else            {                left = mid+1;            }        }        return -1;    }};