LeetCode : Partition List

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.

题意：给一个单链表和一个值x，将单链表以x为基准进行划分，左边小于x，右边大于等于x。（注意：不能改变初始相对顺序）

解题思路：定义head1和head2，遍历链表，将小于x的节点链接到head1后面，将大于等于x的节点连接到head2的后面，最后将head2链接到head1的末尾，返回head1就OK

ListNode *partition(ListNode *head, int x) 
{
	if (NULL == head)
		return NULL;

	ListNode head1(0), head2(0);
	ListNode* p1 = &head1, *p2 = &head2;
	
	while (head)
	{
		if (head->val < x)
		{
			p1->next = head;
			p1 = p1->next;
		}
		else
		{
			p2->next = head;
			p2 = p2->next;
		}
		head = head->next;
	}

	p2->next = NULL;
	p1->next = head2.next;
	return head1.next;
}