题目
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.
题意：给定一个单链表和一个x，把链表中小于x的放到前面，大于等于x的放到后面，每部分元素的原始相对位置不变。
思路：其实很简单，遍历一遍链表，把小于x的都挂到head1后，把大于等于x的都放到head2后，最后再把大于等于的链表挂到小于链表的后面就可以了。
实现代码：
/**
* 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) {
ListNode* lessHead = new ListNode(0);
ListNode* greaterHead = new ListNode(0);
ListNode* node = head, *less = lessHead, *greater = greaterHead;
while (node != NULL)
{
ListNode* next=node->next; //保存下一个结点；
if (node->val < x)
{
less->next = node;
less = less->next;
less->next = NULL;
}
else
{
greater->next = node;
greater = greater->next;
greater->next = NULL;
}
node = next; //遍历下一个结点；
}
less->next = greaterHead->next;
return lessHead->next;
}
};