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.
小于x的放前面，大于等于x的放后面
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution { //新建两个链表，一个放小于，一个放大于等于
public:
ListNode* partition(ListNode* head, int x) {
ListNode* head1=new ListNode(0); //这样不用在while里面判断是否为头节点
ListNode* head2=new ListNode(0);
ListNode* l1=head1;
ListNode* l2=head2;
ListNode* p=head;
while(p!=NULL)
{
if(p->val<x)
{
l1->next=p;
l1=l1->next;
}
else
{
l2->next=p;
l2=l2->next;
}
p=p->next;
}
l2->next=NULL;
l1->next=head2->next;
return head1->next;
}
};
原文：http://blog.csdn.net/u011391629/article/details/52142802