Search for a Range [LeetCode]

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Summary: Just a practice of binary search.

 1     vector<int> searchRange(int A[], int n, int target) {

 2         vector<int> range(2,-1);

 3         int start = 0;

 4         int end = n - 1;

 5         while(start <= end) {

 6             int median = start + (end - start + 1) / 2;

 7             if(A[median] > target) {

 8                 end = median - 1;

 9             }else if (A[median] < target) {

10                 start = median + 1;

11             }else { //equals

12                 //go right

13                 int i = median + 1;

14                 for(; i <= end; i ++) {

15                     if(A[i] != target){

16                         range[1] = i - 1;

17                         break;

18                     }

19                 }

20                 if(range[1] == -1)

21                     range[1] = i - 1;

22                 //go left

23                 i = median - 1;

24                 for(; i >= start; i --) {

25                     if(A[i] != target){

26                         range[0] = i + 1;

27                         break;

28                     }

29                 }

30                 if(range[0] == -1)

31                     range[0] = i + 1;

32                 

33                 break;

34             }

35         }

36         return range;

37     }