LeetCode-find the starting and ending position&&find index if the target is found

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].

給一個排序的數組，找出目标值的範圍，若目标值不存在，傳回[-1,-1].

代碼如下：

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target)
    {
        int begin = 0;
        int end = n-1;
        vector<int> vs;
        while (begin <= end)
        {
            int mid = (begin + end) / 2;
            if (A[mid] == target)
            {
                int min = mid;
                while (min >= 0 && A[mid] == A[min])
                    min--;
                int max = mid;
                while (max <= end && A[mid] == A[max])
                {
                    max++;
                }
                vs.push_back(min+1);
                vs.push_back(max - 1);
                return vs;
            }
            else if (A[mid] > target)
            {
                end = mid - 1;
            }
            else
            {
                begin = mid + 1;
            }
        }
        vs.push_back(-1);
        vs.push_back(-1);
        return vs;
    }
};      

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

class Solution {
public:
    int searchInsert(int A[], int n, int target)
    {
        if (A[n-1]< target)
            return n;
        int left = 0, right = n - 1;
        while (left < right)
        {
            int mid = left + (right - left) / 2;
            if (A[mid] == target)
                return mid;
            else if (A[mid] < target)
                left = mid + 1;
            else
                right = mid;
        }
        return right;
    }
};