[LeetCode]86. Partition List

86. Partition List
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：
根据给定链表和给定值X，把大于X值的放在X右边，小于X值的放在X左边。并不改变相对位置，比如4和3都大于等于3，那么他们移到右边后位置仍为4在前，3在后。

思路：
1）链表为空或者只有一个节点，返回即可。
2）定义两个链表list和back，分别存放小于X节点的链表和大于等于X的链表。使用二级指针方便头节点处理。
3）while循环结束后，list和back链表的最后节点的next并未置NULL。所以把back链表尾节点next置NULL。并把back追加到list后面即可。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     struct ListNode *next;
* };
*/
struct ListNode* partition(struct ListNode* head, int x)
{
if ( head == NULL || head->next == NULL )
{
return head;
}
struct ListNode *list = NULL;
struct ListNode **first  = &list;
struct ListNode *back = NULL;
struct ListNode **second = &back;

while ( head )
{
if ( head->val < x)
{
*first = head;
first = &(*first)->next;
}
else
{
*second = head;
second = &(*second)->next;
}

head = head->next;
}

*second = NULL;
*first = back;

return list;
}