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.

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> result;
        if(nums.size() == 0) return result;
        
        int left = findLeft(nums, target);
        int right = findRight(nums, target);
        
        result.push_back(left);
        result.push_back(right);
        
        return result;
    }
    
    int findLeft(vector<int> &nums, int target){
        int low = 0, high = nums.size() - 1;
        
        while(low <= high){
            int mid = (low + high) / 2;
            if(nums[mid] < target) low = mid + 1;
            else high = mid - 1;
        }
        if(nums[low] != target) return -1;
        else return low;
    }
    
    int findRight(vector<int> &nums, int target){
        int low = 0, high = nums.size() - 1;
        
        while(low <= high){
            int mid = (low + high) / 2;
            if(nums[mid] > target) high = mid - 1;
            else low = mid + 1;
        }
        if(nums[high] != target) return -1;
        else return high;
    }
};