问题叙述性说明：

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 ofO(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].

基本思路：

此题能够用二分查找法解决。

假设数组中没有target，能够在O（lgn）时间完毕。

假设数组中有target，当发现target后，对前后的内容继续用二分法 查找这种位置pos 。

1. 前面的pos须要满足 A[pos] < target && A[pos+1] == target. pos+1即是開始位置。
2. 后面的pos须要满足 A[pos] > target && A[pos-1] == target . pos-1是结束位置。

代码：

vector<int> searchRange(int A[], int n, int target) {   //C++
        vector<int> result(2);
        
        int low = 0, high = n-1;
        int mid, begin = -1, end = -1;
        while(low <= high)
        {
            mid = (low+high)/2;
            if(A[mid] > target)
                high = mid - 1;
            else if(A[mid] < target)
                low = mid + 1;
            else 
            {
                    begin = mid;
                    end = mid;
                    
                    //get begin
                    if(low <= begin -1){
                        while((low <= begin-1) && !(A[low]<target && A[low+1] == target) )
                        {
                            mid = (low + begin-1)/2;
                            if(A[mid] < target)
                                low = mid+1;
                            else begin = mid;
                        }
                        if(A[low]<target && A[low+1] == target)
                            begin = low+1;
                    }
                    //get end
                    if(high >= end+1){
                        while((high >= end+1) &&!(A[high]>target && A[high-1] == target))
                        {
                            mid = (high + end +1)/2;
                            if(A[mid] > target)
                                high = mid - 1;
                            else end = mid;
                        }
                        if(A[high]>target && A[high-1] == target)
                            end = high - 1;
                    }
                    break;
                    
            }
        }
        result[0] = begin;
        result[1] = end;
        return result;
    }