面試算法題. 單鏈表反轉的遞歸版本

// 鏈表節點結構
struct ListNode {
	ListNode(int a = 0) : val(a), 
			next(nullptr) {
	}

	int val;
	ListNode* next;
};


// 遞歸版本的鏈表反轉
ListNode* ReverseList(ListNode* head) {

	if (nullptr == head)
		return nullptr;

	if (nullptr == head->next)
		return head;

	ListNode* newHead = ReverseList(head->next);
	head->next->next = head;

	return newHead;
}

int main(int argc, char* argv[]) {


	ListNode* head = new ListNode(1);
	head->next = new ListNode(2);
	head->next->next = new ListNode(3);
	head->next->next->next = new ListNode(4);

	ListNode *newHead = ReverseList(head);

	return 0;
}