Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,

Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

class Solution {public:    ListNode *partition(ListNode *head, int x) {        if(!head || !(head->next) ) return head;        ListNode *current = head;        ListNode *smallPointer = NULL; //point to the last node 
     
      x        while(current)        {            if(current->val >= x)            {                largePointer = current;                current = current->next;            }            else            {                if(!largePointer)                {                    smallPointer = current;                    current = current->next;                }                else if(smallPointer)                {                    largePointer->next = smallPointer->next;                    smallPointer -> next = current;                    current = current->next;                    smallPointer = smallPointer->next;                    smallPointer->next = largePointer->next;                    largePointer->next = current;                }                else //head                {                    smallPointer = current;                    current = current->next;                    smallPointer->next = head;                    head = smallPointer;                    largePointer->next = current;                }             }        }        return head;            }};