＃14 First Position of Target

题目描述：

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?

题目思路：

这题还是传统的二分法解决问题。

Mycode（AC ＝ 78ms）：

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;

long long l = 0, r = array.size() - 1;
while (l + 1 < r) {
int mid = l + (r - l) / 2;
if (array[mid] < target) {
l = mid + 1;
}
else if (array[mid] > target) {
r = mid - 1;
}
else {
r = mid;
}
}

if (array[l] == target || array[r] == target) {
return array[l] == target? l : r;
}
else {
return -1;
}
}
};