Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

 

C++代码实现：

#include<iostream>

#include<new>

using namespace std;



//Definition for singly-linked list.

struct ListNode

{

    int val;

    ListNode *next;

    ListNode(int x) : val(x), next(NULL) {}

};

class Solution

{

public:

    ListNode *deleteDuplicates(ListNode *head)

    {

        if(head==NULL||head->next==NULL)

            return head;

        ListNode *p=head->next;

        ListNode *pre=head;

        ListNode *ppre=head;

        ListNode *q=NULL;

        while(p)

        {

            if(p->val!=pre->val)

            {

                //要删除头结点的情况，如果pre与ppre都指向头结点，则不需要删除，否则将删除头结点到p之间的所有结点，包括头结点

                if(ppre==head&&ppre->val==pre->val&&pre!=ppre)

                {

                    while(ppre!=p)

                    {

                        q=ppre;

                        ppre=ppre->next;

                        q->next=NULL;

                        delete q;

                    }

                    head=p;

                    ppre=p;

                    pre=p;

                    p=p->next;

                    continue;

                }

                else if(ppre->next!=pre)

                {

                    pre->next=NULL;

                    pre=ppre->next;

                    ppre->next=p;

                    while(pre)

                    {

                        q=pre;

                        pre=pre->next;

                        q->next=NULL;

                        delete q;

                    }

                    pre=p;

                    p=p->next;

                    continue;

                }

                ppre=pre;

                pre=p;

                p=p->next;

            }

            else

            {

                pre=p;

                p=p->next;

            }

        }

        cout<<ppre->val<<endl;

        cout<<pre->val<<endl;

        //如果ppre需要删除，肯定是因为指向头结点，否则ppre指向的结点不会与pre相等

        if(ppre==head&&ppre->val==pre->val&&ppre!=pre)

        {

            while(ppre!=p)

            {

                q=ppre;

                ppre=ppre->next;

                q->next=NULL;

                delete q;

            }

            return NULL;

        }

        else if(ppre->next!=pre)

        {

            pre->next=NULL;

            pre=ppre->next;

            ppre->next=p;

            while(pre)

            {

                q=pre;

                pre=pre->next;

                q->next=NULL;

                delete q;

            }

        }

        return head;

    }

    void createList(ListNode *&head)

    {

        ListNode *p=NULL;

        int i=0;

        int arr[10]= {6,5,5,4,4,3,3,3,2,2};

        for(i=0; i<2; i++)

        {

            if(head==NULL)

            {

                head=new ListNode(arr[i]);

                if(head==NULL)

                    return;

            }

            else

            {

                p=new ListNode(arr[i]);

                p->next=head;

                head=p;

            }

        }

    }

};



int main()

{

    Solution s;

    ListNode *L=NULL;

    s.createList(L);

    ListNode *head=L;

    while(head)

    {

        cout<<head->val<<" ";

        head=head->next;

    }

    cout<<endl;

    L=s.deleteDuplicates(L);

    while(L)

    {

        cout<<L->val<<" ";

        L=L->next;

    }

}

    

 分别使用一个指针指向p节点的前驱节点pre，一个指向q节点的前驱节点qre，这样如果p和qre不相等，就删除p到qre之间的节点，注意最后q为空的情况，需要讨论是否要删除节点就是看p和qre是否相等。

    ListNode *deleteDuplicates(ListNode *head) {

        if(head==NULL||head->next==NULL)

            return head;

        ListNode *pre=head;

        ListNode *p=head;

        ListNode *qre=head;

        ListNode *q=p->next;

        while(q)

        {

            if(p->val!=q->val)

            {

                if(p->next==q)

                {

                    pre=p;

                    p=q;

                    qre=q;

                    q=q->next;

                }

                else

                {

                    qre->next=NULL;

                    if(p==head)

                    {

                        while(p)

                        {

                            ListNode *tmp=p->next;

                            delete p;

                            p=tmp;

                        }

                        head=q;

                        pre=q;

                        p=q;

                        qre=q;

                        q=q->next;

                    }

                    else

                    {

                        pre->next=q;

                        while(p)

                        {

                            ListNode *tmp=p->next;

                            delete p;

                            p=tmp;

                        }

                        p=q;

                        qre=q;

                        q=q->next;

                    }

                }

            }

            else

            {

                qre=q;

                q=q->next;

            }

        }

        if(p!=qre)

        {

            pre->next=q;

            if(head==p)

                head=NULL;

            while(p)

            {

                ListNode *tmp=p->next;

                delete p;

                p=tmp;

            }

        }

        return head;

    }