Search for a range

Given an array of integers sorted in ascending order, 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].

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