leetcode86 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.

/**
 * 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 *ans=new ListNode(0);
        ListNode *ansp=ans;
        
        ListNode *p=head;
        while(p)
        {
            if(p->val<x)
            {
                ListNode *temp=new ListNode(p->val);
                ansp->next=temp;
                ansp=ansp->next;
            }
            p=p->next;
        }
        p=head;
        while(p)
        {
            if(p->val>=x)
            {
                ListNode *temp=new ListNode(p->val);
                ansp->next=temp;
                ansp=ansp->next;
            }
            p=p->next;
        }
        return ans->next;
    }
};