LeetCode Q82: 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.

Solution 1 (Non-recursive)

Compare value of current node and next node. Be careful to check if next node is NULL ahead. Also, be careful of some corner case such as:

[1, 1, 2]: Need to create a new head  node in front of the first node.

[1, 1]: Need to create a new head  node in front of the first node.

[1]: Need to create a new head  node in front of the first node.

[1, 2, 2, 3, 3]: when duplicate numbers are one after another, need to double check whether the first non identical number is a new start of next group of duplicate numbers.

	/**
	* 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* newHead=new ListNode(-1);
	newHead->next=head;
	ListNode* pre=newHead;
	ListNode* cur=head;
	bool dup=false;
	while(cur!=NULL){
	dup=false;
	while(cur!=NULL&&cur->next!=NULL){
	if(cur->val==cur->next->val){
	dup=true;
	cur=cur->next;
	}else
	break;
	}
	cur=(dup==true)?cur->next:cur;
	if(cur==NULL){
	pre->next=cur;
	ListNode* tmpHead=newHead->next;
	delete newHead;
	return tmpHead;
	}
	if(cur->next!=NULL){
	if(cur->val==cur->next->val)
	continue;
	}
	pre->next=cur;
	pre=cur;
	if(cur!=NULL)
	cur=cur->next;
	}
	ListNode* tmpHead=newHead->next;
	delete newHead;
	return tmpHead;
	}
	};

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Round 2 solution:

	/**
	* 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)
	return NULL;
	ListNode* newHead = new ListNode(head->val+1);
	newHead->next = head;
	ListNode* p=newHead;
	ListNode* n=head;
	ListNode* nn=n->next;
	while(nn!=NULL){
	if(nn->val!=n->val){
	p=p->next;
	n=p->next;
	nn=n->next;
	}else{
	while(nn!=NULL&&n->val==nn->val){
	n=n->next;
	nn=n->next;
	}
	p->next=nn;
	n=p->next;
	nn=n==NULL? NULL:n->next;
	}
	}
	return newHead->next;
	}
	};

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