题目：

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].

思路：

遍历一次，直接找出起始和结束位置即可。

代码：

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int start = -1;
        int end = -1;
        
        for(int i = 0; i < n; i++)
        {
            if(A[i] == target && start == -1)
            {
                start = i;
                end = i;
            }
            else if(start != -1 && A[i] == target)
            {
                end += 1;
            }
        }
        
        vector<int> v;
        v.push_back(start);
        v.push_back(end);
        
        return v;
    }
};