82. 删除排序链表中的重复元素 II
题目描述
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.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
方法1
不使用辅助头结点的解决方案
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if(head == nullptr)
return nullptr;
ListNode * ret = nullptr;
ListNode * last = nullptr;//虚尾指针的使用
ListNode * p = head;//当前要判断的节点
while(p!=nullptr)//需要思考：当前节点满足什么条件，才算合格
{
if(p->next==nullptr || p->val!=(p->next)->val)
{
if(last == nullptr)
ret = p;
last = p;
p = p->next;
}
else
{
int val = p->val;
ListNode *next = p;//寻找下一个”候选的“节点
while(next!=nullptr && next->val == val)
{
ListNode *temp = next;
next = next->next;
temp->next = nullptr;
delete temp;
}
if(last!=nullptr)
last->next = next;//这一步，保证不需要特别的操作去连接
p = next;
}
}
return ret;
}
};
方法2
上面思路的使用辅助头结点的解决方案
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if(head == nullptr)
return nullptr;
ListNode *dummy = new ListNode(-1);
dummy->next = head;//这里是直接连好了
ListNode * last = dummy;
ListNode * p = head;
while(p!=nullptr)//变量p的含义：当前要试探的结点
{
if(p->next==nullptr || p->val!=(p->next)->val)
{
last = p;
p = p->next;
}
else
{
int val = p->val;
ListNode *next = p;//next变量的含义：下一轮要试探的结点
while(next!=nullptr && next->val == val)
{
ListNode *temp = next;
next = next->next;
temp->next = nullptr;
delete temp;
}
last->next = next;
p = next;
}
}
ListNode *ret = dummy->next;
dummy->next = nullptr;
delete dummy;
return ret ;
}
};
方法3
递归解决方案
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head)
{
if(head==nullptr || (head->next)==nullptr)
return head;
if (head->val != head->next->val)
{ //巧妙的利用函数返回值携带的信息进行自动的连接，
//而不像循环解决方案中的手动连接
head->next = deleteDuplicates(head->next);
return head;
}
else
{
int val = head->val;
ListNode *next = head;
while (next!=nullptr && next->val==val)
{
ListNode *temp = next;
next = next->next;
temp->next = nullptr;
delete temp;
}
return deleteDuplicates(next);
}
}
};
