LeetCdoe Remove Duplicates from Sorted List II移掉重复链表中的元素

Remove Duplicates from Sorted List II

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

For example,
Given1->2->3->3->4->4->5, return1->2->5.
Given1->1->1->2->3, return2->3.

是重复的元素就全部移掉，前面有一道是移掉多余的重复元素，保留一个。

这道题的关键就需要注意保存重复元素的前一个指针，这样才能移除元素。

考指针和链表的操作熟练程度。

/**
 * 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 || !head->next) return head;
		ListNode dummy(-1);	
		dummy.next = head;
		ListNode *pre = &dummy;
		ListNode *cur = head;
		ListNode *post = head->next;
		bool flag = false;

		while (cur && post)
		{
			while (post && cur->val == post->val)
			{
				post = post->next;
				flag = true;
			}
			if (flag == true) pre->next = post;
			else pre = pre->next;

			flag = false;
			cur = post;
			if (post) post = post->next;
		}
		//不能用return head， 因为如果head重复的话，需要删掉head，但是其实head还在内存中，没有删掉，不过改变了dummy->next的链表。
		return dummy.next;
	}
};

//2014-2-13 update
	ListNode *deleteDuplicates(ListNode *head) 
	{
		if (!head || !head->next) return head;
		ListNode dummy(0);
		ListNode *p = &dummy;
		bool repeat = false;
		for ( ;head->next; head = head->next)//记得步进head
		{
			if (head->val == head->next->val) repeat = true;
			else 
			{
				if (repeat) repeat = false;
				else//不重复的才加入新的链表中
				{
					p->next = head;
					p = p->next;
				}
			}
		}
		if (!repeat)//注意别漏掉
		{
			p->next = head;
			p = p->next;
		}
		p->next = nullptr; //注意：删除结尾多余的节点。
		return dummy.next;
	}