[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:

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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:

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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++：

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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

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