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：给定一个已经排序的数组，求出target第一次出现的位置和最后出现的位置
 //二分法思想。分别找到target第一次出现的位置index1，和target最后出现的位置index2
 //利用(target+1)第一次出现的位置减1，得到target最后出现的位置
 class Solution {
 public:
	 vector<int> searchRange(vector<int>& nums, int target) {
		 int index1 = lower_bound(nums, target);
		 int index2 = lower_bound(nums, target + 1) - 1; //利用(target+1)第一次出现的位置减1，得到target最后出现的位置
		 if (index1 < nums.size() && nums[index1] == target)
			 return{ index1, index2 };
		 else
			 return{ -1, -1 };
	 }

	 int lower_bound(vector<int>& nums, int target) {
		 int left = 0, right = nums.size() - 1;
		 while (left <= right) 
		 {
			 int mid = (right + left) / 2;
			 if (nums[mid] < target)
				 left = mid + 1;
			 else
				 right = mid - 1;
		 }
		 return left;
	 }
 };