LeetCode82 Remove Duplicates from Sorted List II

题目：

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

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

分析：

链表去重的follow-up，但是题号这个在前面，要把有重复元素，则所有该元素均删掉。

注意：

1. 使用dummy node处理头结点被删除掉的情况。

2. 只要有head-> next -> val存在时，先判断head -> next是否为空。

代码：

/**
 * 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) {
        ListNode dummy(0);
        dummy.next = head;
        head = &dummy;
        while(head -> next != nullptr && head -> next -> next != nullptr) {
            if (head -> next -> val == head -> next -> next -> val) {
                int val = head -> next -> val;
                while (head -> next != nullptr && head -> next -> val == val) { //只要取head->next，先判断
                    ListNode* temp = head -> next;
                    head -> next = head -> next -> next;
                    delete temp;
                }
            }
            else {
                head = head -> next;
            }
        }
        return dummy.next;
    }
};