【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.

 

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        if(head==NULL) return NULL;

        ListNode * less=new ListNode(0);
        ListNode * greater=new ListNode(0);
        ListNode * greaterhead=greater;
        ListNode * lesshead=less;
        ListNode * p=head;

        while(p)
        {
            if(p->val<x)
            {
                less->next=p;
                p=p->next;
                less=less->next;
                less->next=NULL;
            }
            else
            {
                greater->next=p;
                p=p->next;
                greater=greater->next;
                greater->next=NULL;
            }
        }

        less->next=greaterhead->next;
        return lesshead->next;
    }
};