【LeetCode】Partition 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.

/**
 * 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) {
        ListNode *smallhead = NULL;
	ListNode *bighead = NULL;
	ListNode *tail_small = NULL;
	ListNode *tail_big = NULL;
	ListNode *cur = head;
	if (head == NULL || head->next == NULL)
	{
		return head;
	}
	while(cur != NULL)
	{
		ListNode *tmp_next = cur->next;
		if (cur->val < x)
		{
			if (smallhead == NULL)
			{
				smallhead = tail_small = cur;
			}
			else {
				tail_small->next = cur;
				tail_small = cur;
			}
		}
		else{
			if (bighead == NULL)
			{
				bighead = tail_big = cur;
			}
			else{
				tail_big->next = cur;
				tail_big = cur;
			}
		}
		cur = tmp_next;
	}
	if (tail_small == NULL)
	{
		tail_big->next = NULL;
		return bighead;
	}
	if (tail_big == NULL)
	{
		tail_small->next = NULL;
		return smallhead;
	}
	tail_small->next = bighead;
	tail_big->next = NULL;
	return smallhead;
    }
};