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     }