leetcode 82. 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.

这道题很简答，遇到这种指针操作的题，最重要的就是细心。

代码如下：

/* class ListNode {      int val;      ListNode next;      ListNode(int x) { val = x; } }*/public class Solution {    public ListNode deleteDuplicates(ListNode head)     {        if(head==null || head.next==null)            return head;        ListNode finalListNode=new ListNode(0);        ListNode resIter=finalListNode;        ListNode iter=head;        while(iter!=null)        {            ListNode fa=iter;            ListNode son=iter.next;                     //son为null表示列表走到尽头            //fa的value和son的value第一次就不等，表示fa可以直接加入res列表            if(son==null|| fa.val!=son.val)            {                resIter.next=fa;                resIter=resIter.next;                resIter.next=null;              }else            {                while(son!=null && fa.val==son.val)                {                    fa=son;                    son=son.next;                }            }            iter=son;        }        return finalListNode.next;    }}

下面是C++的做法，

代码如下：

#include <iostream>#include <vector>using namespace std;/*struct ListNode{    int val;    ListNode *next;    ListNode(int x) : val(x), next(NULL) {}};*/class Solution {public:    ListNode* deleteDuplicates(ListNode* h)     {        if (h == NULL || h->next == NULL)            return h;        ListNode* head = new ListNode(h->val - 2);        head->next = h;        ListNode* i = head;        while (i != NULL)        {            ListNode* fa = i->next;            ListNode* son = fa==NULL?NULL:fa->next;            if (son==NULL || fa->val != son->val)                i = i->next;            else            {                ListNode* j = fa;                while (j != NULL && j->val == fa->val)                    j = j->next;                i->next = j;            }        }        return head->next;    }};