leetcode---partition-list---链表

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,
Given1->4->3->2->5->2and x = 3,
return1->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) 
    {
        if(!head)
            return head;

        ListNode *p = head;
        ListNode *low = new ListNode(0);
        ListNode *high = new ListNode(0);
        ListNode *low1 = low;
        ListNode *high1 = high;

        while(p)
        {
            if(p->val < x)
            {
                low->next = p;
                low = low->next;
            }
            else
            {
                high->next = p;
                high = high->next;
            }
            p = p->next;
        }

        low->next = high1->next;
        high->next = NULL;
        return low1->next;
    }
};