The single list of elements rearranged, the left small, equal in the middle, on the right large (optimization -C ++)

/*
* @Author: hzf
* @Date:   2020-03-01 22:24:23
* @Last Modified by:   hzf
* @Last Modified time: 2020-03-01 22:59:53
*/
/*
   一般方法：
    在数组中进行快排，然后放入单链表中
*/

/*优化算法
 通过六个变量实现单链表中小于num的数字在左边，等于在中间，大于的在右边
 实现方法:
 建立六个指针分别指向小于区的首尾
 				   等于区的首尾
 				   大于区的首尾
 	然后对原链表进行遍历，分别放在相应的首尾指针中
 	最后分类讨论是否有空指针，并且把首尾指针串起来
*/
#include <iostream>
using namespace std;
class Node
{
public:
	Node(int data)
	{
		this.value = dat ;
	}
	int value;
	Node* next;
};



Node* listPartitionq(Linklist head, int pivot)
{
	Node sH = NULL;//small head
	Node sT = NULL;//small tail
	Node eH = NULL;//equal head
	Node eT = NULL;//equal tail
	Node bH = NULL;//big head
	Node bT = NULL;//big tail
	Node next = NULL;//save next node

	while(head != NULL){
		next = head.next;
		head.next = NULL;

		if (head.value < pivot)
		{
			if (sH == NULL)
			{
				sH = head;
				sT = head;
			}else{
				sH.next = head
				sT = head;
			}
			else if (head.value == pivot)
			{
				if (eH == NULL)
				{
					eH = head;
					eT = head;
				}else{
					eH.next = head;
					eT = head;
				}
			}
			else{
				if (bH == NULL)
				{
					bH = head;
					bT = head
				}else{
					bH.next = head;
					bT = head;
				}
			}
		}
	}
	head = next;

	//small and equal reconnect
		if (sT != null) {
			sT.next = eH;
			eT = eT == null ? sT : eT;
		}
		// all reconnect
		if (eT != null) {
			eT.next = bH;
		}
		return sH != null ? sH : eH != null ? eH : bH;
}

int main(int argc, char const *argv[])
{
	/* code */
	return 0;
}