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.

解题思路:对于小于x的结点重新构成一个链表，删除原链表的该值，最后连接连个链表;

#include<iostream>
#include<vector>
using namespace std;
//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
ListNode *partition(ListNode *head, int x) {
	ListNode*TmpHead   = head;
	ListNode*PreNode   = head;
	ListNode*LessXList = new ListNode(0);
	ListNode*TmpLessXhead = LessXList;
	while (TmpHead!=NULL){
		if (TmpHead->val<x)
		{
			TmpLessXhead->next = TmpHead;
			TmpLessXhead = TmpLessXhead->next;
			if (TmpHead == head)
				head = head->next;
			else
				PreNode->next = TmpHead->next;
		}
		else
			PreNode = TmpHead;
		TmpHead = TmpHead->next;
	}
	TmpLessXhead->next = head;
	return LessXList->next;
}

Partition List