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.

/** * 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 *left = new ListNode(0);        ListNode *right = new ListNode(0);        ListNode *l = left;        ListNode *r = right;        ListNode *p = head;        if(head == NULL)            return NULL;        while(p)        {            if(p->val < x)            {                left->next = p;                left = left->next;            }            else            {                right->next = p;                right = right->next;            }            p = p->next;        }        right->next = NULL;        left->next = r->next;        return l->next;    }};