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].
不知道这样行不行，general上是O（log n）了，只是在某些bad case时会退化为O(n).
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 l = 0;
int r = n - 1;
int m = 0;
vector<int> res(2);
res[0] = -1;
res[1] = -1;
while(l <= r){
m = (l + r)/2;
if(A[m] =&#
