[LeetCode] 82. Remove Duplicates from Sorted List II_Medium tag: Linked List

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinctnumbers from the original list.

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

Input: 1->1->1->2->3
Output: 2->3

这个题目跟[LeetCode] 83. Remove Duplicates from Sorted List_Easy tag: Linked List很类似，但是有可能要把head都去掉，所以还是要加上dummy node，并且使得dummy.next = head， 同时设置pre = dummy instead of None, 另外while loop判断条件为head and head.next, 最后返回dummy.next.

Code

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def removeDup2(self, head):
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy
        while (pre.next and pre.next.next):
            if pre.next.val == pre.next.next.val:
                val = pre.next.val
                while (pre.next and pre.next.val == val): # 把重复的点都删掉
                    pre.next = pre.next.next
            else:
                pre = pre.next
        return dummy.next

2)

class Solution:
    def removeDup2(self, head):
        dummy = ListNode(0)
        dummy.next = head
        pre = dummy
        while head and head.next:
            if head.val == head.next.val:
                val = head.val
                while head and head.val == val:
                    pre = head.next
                    head.next = None
                    head = pre.next
            else:
                pre = head
                head = head.next
        return dummy.next