[LeetCode] Search for a Range(二分法)

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].

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        set<int> sIndex;
        search(A,,n-,target,sIndex);
        vector<int> result(,-);
        if(!sIndex.empty()){
            result[] = *sIndex.begin();
            result[] = *(--sIndex.end());
        }
        return result;

    }
private:
    void search(int A[],int start,int end,int target,set<int> &sIndex){

        if(start == end){
            if(A[start]==target){
               sIndex.insert(start);
               return;
            }else
               return;
        }

        int startIndex,endIndex;
        int mid = start + (end-start)/;

        if(A[start]<=target && A[mid]>= target)
             search(A,start,mid,target,sIndex);
        if(A[mid+]<=target && A[end]>= target)
             search(A,mid+,end,target,sIndex);
    }//end func
};