Description: Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Link: https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

Examples:

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

思路: 和83的區別在於，1) 只要這個元素出現過兩次及以上，就刪除，所以當前元素不僅要和前面的比較，還要和後面的比較，當與前後都不同時，才被新增到返回連結串列中，2) pre的更新也不同，83中，當pre和當前值不同時，才更新，相同則不更新，這裡要隨著指標移動而更新。3) 注意return linked list的head，如果從連結串列的第二個元素開始遍歷，需要先確認第一個元素是否unique.

# Definition for singly-linked list.
 

# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head: return head
        
 
if not head.next: return head
        
        rhead = ListNode(0)
        r = rhead
        pre = head.val
        p = head.next
        if pre != p.val:
            r.next = head
            r = r.next
        while p:
            if p.val != pre and (p.next is None or p.val != p.next.val):
                r.next = p
                r = r.next
            pre = p.val
            p = p.next
        r.next = None
        return rhead.next           

日期: 2020-11-17 週二就又過去了半天