[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.

Solution:

/**

 * 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 *ans_head = NULL,*ans_end = NULL, *tmp = head;

        while(tmp != NULL)

        {

            if(tmp -> val < x) 

            {   

                ListNode *ln = new ListNode(tmp -> val);

                if(ans_head == NULL)

                {

                    ans_head = ln;

                    ans_end = ln;

                }

                else

                {

                    ans_end -> next = ln;

                    ans_end = ln;

                }

            }

            

            tmp = tmp -> next;

        }

        

        tmp = head;

        while(tmp != NULL)

        {

            if(tmp -> val >= x) 

            {   

                ListNode *ln = new ListNode(tmp -> val);

                if(ans_head == NULL)

                {

                    ans_head = ln;

                    ans_end = ln;

                }

                else

                {

                    ans_end -> next = ln;

                    ans_end = ln;

                }

            }

            

            tmp = tmp -> next;

        }

        

        return ans_head;

    }

};