Given a linked list, rotate the list to the right by k places, where k is non-negative.
Example 1:
Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL
Example 2:
Input: 0->1->2->NULL, k = 4 Output:2->0->1->NULLExplanation: rotate 1 steps to the right: 2->0->1->NULL rotate 2 steps to the right: 1->2->0->NULL rotate 3 steps to the right:0->1->2->NULLrotate 4 steps to the right:2->0->1->NULL
很特殊的一点是采用先构成环，改变开头再断开的方式
# Definition for singly-linked list. class ListNode(object): def __init__(self, x): self.val = x self.next = None class Solution(object): def rotateRight(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode """ if k==0:return head if head==None:return head dummy=ListNode(0) dummy.next=head p=dummy count=0 while p.next: p=p.next count+=1 # 这里成环了，p和dummy是相等的，dummy代表的是开始的数值，p已经走到了结尾，结尾又连到了开始 p.next=dummy.next step=count-(k%count) for i in range(0,step): p=p.next head=p.next # 在这步之前都是成环的，输出head会不停的，之前形成环，算step在合适的地方剪断 p.next=None # while True: # print("start") # print(head.val) # if not head.next: break # head = head.next return head solu=Solution() head=ListNode(1) l2=ListNode(2) l3=ListNode(3) l4=ListNode(4) l5=ListNode(5) head.next=l2 l2.next=l3 l3.next=l4 l4.next=l5 k=2 ans=solu.rotateRight(head,k) while True: print(ans.val) if not ans.next:break ans=ans.next
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers 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
import collections # from collections import Counter # Definition for singly-linked list. class ListNode(object): def __init__(self, x): self.val = x self.next = None class Solution(object): def deleteDuplicates(self, head): """ :type head: ListNode :rtype: ListNode """ root=ListNode(0) root.next=head val_list=[] # 形成字典 while head: val_list.append(head.val) head=head.next counter=collections.Counter(val_list) head=root while head.next: if counter[head.next.val]!=1: head.next=head.next.next else: head=head.next return root.next solu=Solution() head=ListNode(1) l2=ListNode(1) l3=ListNode(2) l4=ListNode(2) l5=ListNode(4) head.next=l2 l2.next=l3 l3.next=l4 l4.next=l5 ans=solu.deleteDuplicates(head) while True: print(ans.val) if not ans.next: break ans=ans.next
LeetCode开心刷题四十四天——61. Rotate List 82. Remove Duplicates from Sorted List II(时间慢，空间70)
原文：https://www.cnblogs.com/Marigolci/p/11547209.html