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.

     vector<int> searchRange(int A[], int n, int target) {
         vector<int> range(,-);
         int start = ;
         int end = n - ;
         while(start <= end) {
             int median = start + (end - start + ) / ;
             if(A[median] > target) {
                 end = median - ;
             }else if (A[median] < target) {
                 start = median + ;
             }else { //equals
                 //go right
                 int i = median + ;
                 for(; i <= end; i ++) {
                     if(A[i] != target){
                         range[] = i - ;
                         break;
                     }
                 }
                 if(range[] == -)
                     range[] = i - ;
                 //go left
                 i = median - ;
                 for(; i >= start; i --) {
                     if(A[i] != target){
                         range[] = i + ;
                         break;
                     }
                 }
                 if(range[] == -)
                     range[] = i + ;

                 break;
             }
         }
         return range;
     }