Leetcode 61

Posted on | | 阅读数
Words count in article: 196 | Reading time ≈ 1

Question

Solution

首先需要考虑链表的总长度，因为需要用它来计算执行了多少个循环。通过k%length来判断实际需要执行的部分。创建两个指针，然后将指向末位元素的指针的next指向head，将前面的指针next设为None。这个时候，需要注意的是，head已经不是整个链表的head了，需要调整。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        if not head:
            return 
        if head.next == None:
            return head
        if k == 0:
            return head
        pointer = head
        length = 1
        while pointer.next:
            pointer = pointer.next
            length += 1
        
        roate = k % length
        if roate == 0:
            return head
        pre = cur = head
        while cur.next:
            if roate:
                roate -= 1
            else:
                pre = pre.next
            cur = cur.next
        temp = pre.next
        pre.next = None
        cur.next = head
        head = temp
        return head