Remove Duplicates from Sorted List II

Given a sorted linked list, delete all duplicates such that each element appear only once.

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

这个题目比较简单，用两个指针来完成。后面的指针与前面的重复则跳过。注意链表尾结点元素的处理。

/**
* 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 || head->next==NULL)
return head;
ListNode *p=head,*q=p->next;
while(q!=NULL){
if(p->val==q->val && q->next==NULL){ //相等的节点正好为尾节点，则将p->next置为0，返回即可
p->next=NULL;
break;
}
if(p->val==q->val)
q=q->next;
else{
p->next=q;
p=p->next;
q=q->next;
}
}
return head;
}
};

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-&
4000
gt;5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

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这个题我用一次遍历来做。p和q来判断是否出现重复的元素。tmp记录p的前一个有效的节点。题目中的例子都可以通过，但是[1,1]的输入会出现Runtime Error，暂时还没有找到原因，代码如下；

/**
* 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 || head->next==NULL)
return head;

ListNode *p=head, *q=p->next,*tmp=p;
bool flag=false;
while(q!=NULL){
while(p->val==q->val){
p=q;
q=q->next;
flag=true;
}
if(flag){
if(head==tmp){
head=q;
tmp=q;
}
p=q;
q=q->next;
flag=false;
}
else{
tmp=p;
p=q;
q=q->next;
}
}
return head;
}
};

后来又学习了递归的解法，代码比较少，可以很轻松的完成要求。代码如下：

/**
* 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 || head->next==NULL)
return head;

ListNode *p=head;
while(p->next!=NULL && p->val==p->next->val)
p=p->next;
if(p!=head){
while(head!=p->next){
ListNode *tmp=head;
head=head->next;
free(tmp);
}
return deleteDuplicates(p->next);
}
head->next=deleteDuplicates(head->next);
return head;
}
};