【leetcode】Partition List

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.
思路：
用两个指针，分别指向比x小的数，和其他的数，然后再整体地连起来，就OK了。注意最后得时候，要指向NULL.

/** * 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 || head->next==NULL) return head;        ListNode *p, *q,*temp=NULL,*record;        ListNode *newnode=(struct ListNode *)malloc(sizeof(struct ListNode));        ListNode *oldnode=(struct ListNode *)malloc(sizeof(struct ListNode));        record=oldnode;        q=newnode;        p=head;        while(p)        {            if(p->val >= x)            {                q->next=p;                q=q->next;            }            else            {                record->next=p;                record=record->next;            }            p=p->next;        }        q->next=NULL;        record->next=NULL;        record->next=newnode->next;        return oldnode->next;    }};