86. Partition List (List)

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
        ListNode *largePointer = 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;

    }
};