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].

这道题可以先用二分查找找到target所在的位置location, 然后以location为中心向左右两边扩展，直到值与target不同为止。但这种方法最坏情况的复杂度是O(n)，我们可以考虑数组中全是相同元素的情况。奇怪的是代码还是被accepted。此处binary search用迭代法实现：

 

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> result;
        
        int l = 0, r = n-1;
        while(l <= r){
            int mid = (l + r)/2;
            if(target == A[mid]){
                int low = mid, high = mid;
                while(low >= 0 && A[low] == target)
                    low--;
                while(high < n && A[high] == target)
                    high++;
                result.push_back(low+1);
                result.push_back(high-1);
                return result;
            }else if(A[mid] < target){
                l = mid + 1;
            }else{
                r = mid - 1;    
            }
        }
        
        return vector<int>(2,-1);
    }
};