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.
思路：先分成大于等于x 和 小于x 两个链表 再连起来 还是用伪头部
class Solution {
public:
ListNode *partition(ListNode *head, int x) {
ListNode large(), small();
ListNode * l = &large;
ListNode * s = &small;
while(head != NULL)
{
if(head->val < x)
{
s = s->next = head;
}
else
{
l = l->next = head;
}
head = head->next;
}
l->next = NULL;
s->next = large.next;
return small.next;
}
};