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的区别在于，1) 只要这个元素出现过两次及以上，就删除，所以当前元素不仅要和前面的比较，还要和后面的比较，当与前后都不同时，才被添加到返回链表中，2) pre的更新也不同，83中，当pre和当前值不同时，才更新，相同则不更新，这里要随着指针移动而更新。3) 注意return linked list的head，如果从链表的第二个元素开始遍历，需要先确认第一个元素是否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 周二就又过去了半天