给定一个排序链表,删除所有含有重复数字的节点,只保留原始链表中 没有重复出现 的数字。
示例 1:
输入: 1->2->3->3->4->4->5
输出: 1->2->5
示例 2:
输入: 1->1->1->2->3
输出: 2->3
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii
递归解法:
/**
* 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* p = head->next;
if(head->val == p->val) //把有重复的全删了,返回deleteDuplicates(p)
{
while(p && p->val == head->val)
{
ListNode* tmp = p;
p = p->next;
delete tmp;
}
delete head;
return deleteDuplicates(p);
}
else
{
head->next = deleteDuplicates(head->next);
return head;
}
}
};
迭代:
/**
* 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 == nullptr)
return head;
ListNode dummy(INT_MIN); // 头结点
dummy.next = head;
ListNode *prev = &dummy, *cur = head;
while (cur != nullptr)
{
bool duplicated = false;
while (cur->next != nullptr && cur->val == cur->next->val)
{
duplicated = true;
ListNode *temp = cur;
cur = cur->next;
delete temp;
}
if (duplicated)
{
ListNode *temp = cur;
cur = cur->next;
delete temp;
continue;
}
prev->next = cur;
prev = prev->next;
cur = cur->next;
}
prev->next = cur;
return dummy.next;
}
};