Partition List

Problem:
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.

Solution:
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *partition(ListNode *head, int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        ListNode* small = NULL, *smallHead = NULL, *temp, *bigHead = NULL;
        temp = head;
       
        while (temp != NULL && (temp->val < x)) {
            if (small == NULL) {
                smallHead = small = temp;
            } else {
                small->next = temp;
                small = small->next;
            }
            temp = temp->next;
            // making small->next = NULL is optional
            small->next = NULL;
        }
        bigHead = temp;
        while (temp != NULL && temp->next != NULL) {
            if (temp->next->val < x){
                if (small == NULL) {
                    smallHead = small = temp->next;
                } else {
                    small->next = temp->next;
                    small = small->next;
                }
                temp->next = temp->next->next;
                // making small->next = NULL is optional
                small->next = NULL;
                continue;
            }
            temp = temp->next;
        }
        if (small) {
            small->next = bigHead;
            return smallHead;
        }
        return head;
    }
};