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.

思路：划分链表结构，大于等于目标值移到链表尾部，并更新链表，小于目标值则不动。此题使用使用四个指针来定位链表结构，pPre指向该节点的前一个结点，pLeft代表当前结点，pRight代表移动前的链表尾指针，tail代表时刻更新的链表尾部。首先pLeft向尾部移动，如果该节点值大于等于目标值，则将该节点移到链表尾部，更新pPre和tail，反之pPre=pLeft,pLeft=pLeft->next;最后要注意的是pRight移动前的链表尾结点，该值是否大于等于目标值，如果是，在进行移动，如果不是，没有作为。

/**

 * 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==NULL)

            return NULL;

        ListNode *pPre=NULL;

        ListNode *pLeft=head;

        ListNode *pRight=head;

        while(pRight->next)

        {

            pRight=pRight->next;

        }

        ListNode *pTail=pRight;

        while(pLeft!=pRight)

        {

            if(pLeft->val>=x)

            {

                ListNode *pNext=pLeft->next;

                if(head==pLeft)

                    head=pNext;

                pTail->next=pLeft;

                pLeft->next=NULL;

                pTail=pLeft;

                if(pPre)

                    pPre->next=pNext;

                pLeft=pNext;

            }

            else

            {

                pPre=pLeft;

                pLeft=pLeft->next;

            }

        }

        if(pLeft->val>=x && pRight!=pTail)

        {

            ListNode *pNext=pLeft->next;

            if(head==pLeft)

                head=pNext;

            pTail->next=pLeft;

            pLeft->next=NULL;

            if(pPre)

                pPre->next=pNext;

        }

        return head;

    }

};