[LeetCode]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.

思考：遍历链表，小于x的存于链表head1，大于等于的存于链表head2，合并head1，head2.

/**

 * 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 head;

        ListNode *p=head;

        ListNode *head1=NULL;

        ListNode *head2=NULL;

        ListNode *p1,*p2;

        while(p)

        {

            ListNode *node=new ListNode(p->val);

            if(node->val<x) 

            {

                if(head1==NULL) 

                {

                    head1=p1=node;

                }

                else

                {

                    p1->next=node;

                    p1=node;

                }

            }

            else

            {

                if(head2==NULL) 

                {

                    head2=p2=node;

                }

                else

                {

                    p2->next=node;

                    p2=node;

                }

            }

            p=p->next;

        }

        p1->next=head2;

        if(head1==NULL) return head2;

        return head1;

    }

};