[LeetCode]Search for a range

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].

题目要求：查找有序数组，找到target的范围，如果target不存在于数组，返回[-1, -1]。

思路：和二分查找的思路类似，找到target之后，将start和end同时指向mid，终止查找。然后start向左搜索，end向右搜索，定位区间。

注意start和end下标不要越界的问题(第一次WA就是因为这个错误)。

代码

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        
        vector<int> range;
        if(n==0 || A==NULL)
        {
            range.push_back(-1);
            range.push_back(-1);
            return range;
        }
        
        int l=0, r=n-1, start=-1, end=-1;
        int mid = 0;
        bool success_flag = false;
        
        while(l<=r)
        {
            mid = (l+r)/2;
            if(A[mid] == target)
            {
                start = mid; end = mid;
                success_flag = true;
                break;
            }
            else{
                if(A[mid]>target) // ture left
                {
                    r = mid-1;
                }
                else
                    l = mid+1;
            }
        }
        
        // search range
        if(success_flag)
        {
            while(A[start-1]==target && start>0)
                --start;
            while(A[end+1] == target && end<n-1)
                ++end;
        }
        range.push_back(start);
        range.push_back(end);
        
        return range;
    }
};