OJ lintcode 链表插入排序

用插入排序对链表排序
您在真实的面试中是否遇到过这个题？
Yes
样例
Given 1->3->2->0->null, return 0->1->2->3->null

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @return: The head of linked list.
     */

    //为了化简 ，这是是一个带有头节点的 head节点
    //将node插入道head上面，并且保持从小到大的顺序
    void insert(ListNode *head,ListNode *node){
        node->next=NULL;
        if(head->next==NULL){
            head->next=node;
            return ;
        }
        ListNode * pre=head;
        ListNode * p=head->next;

        while(p!=NULL){
            if(p->val>node->val){
                node->next=p;
                pre->next=node;
                return ;
            }
            else{
                p=p->next;
                pre=pre->next;
            }
            
        }
        pre->next=node;
    }

    ListNode *insertionSortList(ListNode *head) {
        // write your code here
        ListNode * newhead=new ListNode();
        ListNode * p=head;

        while(p!=NULL){
            ListNode * node =p;
            p=p->next;
            insert(newhead,node);
        }
        return newhead->next;
    }
};