Search for a Range

Q: 

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].

A: 二分查找。二分找出target在数组中出现的第一个位置和最后一个位置。

    vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int firstPos = -1;
        int lastPos = -1;
        
        firstPos = searchFirst(A,n,target);
        if(firstPos!=-1)
            lastPos = searchLast(A,n,target);
        
        vector<int> res;
        res.push_back(firstPos);
        res.push_back(lastPos);
        return res;
        
    }
    
    int searchFirst(int A[],int n,int target)
    {
        int b=0,e=n-1;
        int m;
        while(b<e)
        {
            m = b+(e-b)/2;
            if(A[m]<target)
                b = m+1;
            else
                e = m;
        }
        
        if(A[b]==target)
            return b;
        else return -1;
    }
    
    int searchLast(int A[],int n,int target)
    {
        int b=0,e=n-1;
        int m;
        while(b<e)
        {
            m = b+(e-b+1)/2;
            if(A[m]>target)
                e = m-1;
            else
                b = m;
        }
        
        if(A[b]==target)
            return b;
        else return -1;
    }