LintCode 113:Remove Duplicates from Sorted List II

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

Note:

1.由于重复的结点全要删掉，所以使用一个int变量num记录当前判断是否有重复的结点，numNode指向这个结点。当出现重复结点时，若numNode不为空，则删除他的内容，若numNode为空，则只用删除当前重复节点（nextNode)。

2.当head为要删除的结点时，重置head指针为nextNode->next并删除nextNode结点。

3.当当前结点（nextNode)不等于num值时，有两种情况：

 numNode为空，则置numNode=nextNode，更新num值，nextNode后移一格。

 numNode不为空，则preNode，numNode，nextNode全都后移一格，更新num值。

Code:

class Solution {public:    /*     * @param head: head is the head of the linked list     * @return: head of the linked list     */    ListNode * deleteDuplicates(ListNode * head) {        // write your code here        if(!head)            return nullptr;        int num=head->val;        ListNode* preNode=nullptr;        ListNode* numNode=head;        ListNode* nextNode=head->next;        while(nextNode){            if(num==nextNode->val){                if(numNode){                    delete numNode;                    numNode=nullptr;                }                if(preNode){                    preNode->next=nextNode->next;                    delete nextNode;                    nextNode=preNode->next;                }                else{                    head=nextNode->next;                    delete nextNode;                    nextNode=head;                }            }            else{                if(numNode){                    preNode=numNode;                }                numNode=nextNode;                num=numNode->val;                nextNode=nextNode->next;            }        }        return head;    }};