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.
算法思想：
两个指针，s将小于x的串起来，e将大于等于x的串起来，最后将s的next指向e，注意处理head指针就好了
/**
* 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)return NULL;
ListNode *p,*q,*e,*s,*start,*end;
start=s=new ListNode(0);
end=e=new ListNode(0);
p=head;
while(p){
if(p->val<x){
s->next=p;
s=p;
}
else{
e->next=p;
e=p;
}
p=p->next;
}
e->next=NULL;
s->next=end->next;
return start->next;
}
};原文：http://blog.csdn.net/starcuan/article/details/18715789