Leetcode 61. Rotate List

Decription: Given the head of a linked list, rotate the list to the right by k places.

Link: https://leetcode.com/problems/rotate-list/

思路: 假设链表的长度是 length, 通过例子，我们发现 k < length, rotate k times, k >= length, k = k % length times should be rotated. 所以 1) k = k % length. 然后我们发现向右移动k个，实际上是找到倒数第k个位置，把后面的全部元素放到开头来，所以只需找到两个关键Node, 倒数 k-1 (temp)和结尾end，然后第k个元素是new head (nhead)，end.next = head, temp.next = None.

# 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 rotateRight(self, head, k):
        """
        :type head: ListNode
        :type k: int
        :rtype: ListNode
        """
        if not head: return head
        if not head.next: return head
        if k == 0: return head
        
        length = 0
        nodes = []
        p = head 
        while p:
            nodes.append(p)
            p = p.next
            length += 1
        # print(length)
        k = k % length
        if k == 0:
            return head
        else:
            temp = nodes[length - k -1]
            nhead = temp.next
            temp.next = None
            end = nodes[length - 1]
            end.next = head
            return nhead

日期: 2020-11-16 今天regular meeting 被提早了，惊慌失措，又是暖和的冬日