Search for a Range

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].

Analyse: Binary search.

Runtime: 12ms.

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         vector<int> result;
 5         if(nums.size() == 0) return result;
 6         
 7         int left = findLeft(nums, target);
 8         int right = findRight(nums, target);
 9         
10         result.push_back(left);
11         result.push_back(right);
12         
13         return result;
14     }
15     
16     int findLeft(vector<int> &nums, int target){
17         int low = 0, high = nums.size() - 1;
18         
19         while(low <= high){
20             int mid = (low + high) / 2;
21             if(nums[mid] < target) low = mid + 1;
22             else high = mid - 1;
23         }
24         if(nums[low] != target) return -1;
25         else return low;
26     }
27     
28     int findRight(vector<int> &nums, int target){
29         int low = 0, high = nums.size() - 1;
30         
31         while(low <= high){
32             int mid = (low + high) / 2;
33             if(nums[mid] > target) high = mid - 1;
34             else low = mid + 1;
35         }
36         if(nums[high] != target) return -1;
37         else return high;
38     }
39 };