[LeetCode] 82. Remove Duplicates from Sorted List II 移除有序链表中的重复项 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.
和83. Remove Duplicates from Sorted List 类似，这个题是有重复的元素就全部去掉，只保留独立的元素。
解法：去除重复元素的方法和83类似。但由于第一个元素有可能是重复的，如果删除了就不能往下接了，所以新建一个dummy, dummy.next指向链表。最后返回dummy.next就可以了。
Java:
class Solution {
public ListNode deleteDuplicates(ListNode head) {
if(head == null)
return head;
ListNode helper = new ListNode(0);
helper.next = head;
ListNode pre = helper;
ListNode cur = head;
while(cur!=null)
{
while(cur.next!=null && pre.next.val==cur.next.val)
{
cur = cur.next;
}
if(pre.next==cur)
{
pre = pre.next;
}
else
{
pre.next = cur.next;
}
cur = cur.next;
}
return helper.next;
}
}
Python:
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
def __repr__(self):
if self is None:
return "Nil"
else:
return "{} -> {}".format(self.val, repr(self.next))
class Solution(object):
def deleteDuplicates(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
dummy = ListNode(0)
pre, cur = dummy, head
while cur:
if cur.next and cur.next.val == cur.val:
val = cur.val;
while cur and cur.val == val:
cur = cur.next
pre.next = cur
else:
pre.next = cur
pre = cur
cur = cur.next
return dummy.next
C++：
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if (!head || !head->next) return head;
ListNode *start = new ListNode(0);
start->next = head;
ListNode *pre = start;
while (pre->next) {
ListNode *cur = pre->next;
while (cur->next && cur->next->val == cur->val) cur = cur->next;
if (cur != pre->next) pre->next = cur->next;
else pre = pre->next;
}
return start->next;
}
};
类似题目：
[LeetCode] 83. Remove Duplicates from Sorted List 移除有序链表中的重复项
