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.
把小于x的链表单独取出组成新的链表，然后把两个链表相连。
/**
* 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) {
if(head==NULL || head->next==NULL)
return head;
ListNode* p=head;
ListNode* preP=head;
ListNode* newHead=NULL;
ListNode* newP=newHead;
while(p)
{
if(p->val < x)
{
ListNode* tmp=p->next;
if(p==head)
{
head=tmp;
preP=tmp;
}
if(!newHead)
{
newHead=p;
newHead->next=NULL;
newP=newHead;
}
else
{
newP->next=p;
newP=newP->next;
newP->next=NULL;
}
if(preP!=tmp)
preP->next=tmp;
p=tmp;
}
else
{
if(preP!=p)
preP=preP->next;
p=p->next;
}
}
if(!newHead)
return head;
newP->next=head;
return newHead;
}
};