LeetCode 61. 旋转链表

予早2024-07-24 0:07

给你一个链表的头节点 head ，旋转链表，将链表每个节点向右移动 k 个位置。

示例 1：

输入：head = $1,2,3,4,5$ , k = 2

输出： $4,5,1,2,3$

示例 2：

输入：head = $0,1,2$ , k = 4

输出： $2,0,1$

提示：

链表中节点的数目在范围 $0, 500$ 内

-100 <= Node.val <= 100

0 <= k <= 2 * 109

python 复制代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if k == 0 or not head or not head.next:
            return head
        
        _head = head
        total = 0
        while _head:
            total += 1
            _head = _head.next
        _k = k % total

        if _k == 0:
            return head

        dummy = ListNode(next=head)
        slow = fast = dummy
        while fast.next:
            if _k < 1:
                slow = slow.next
            fast = fast.next
            _k -= 1

            
        first = slow.next
        slow.next = None  # forth
        second = fast
        third = dummy.next
        dummy.next = first
        second.next = third
        
        return dummy.next

时间复杂度O(n)

空间复杂度O(1)

看了一眼官解，发现是先连成环然后再断开，比上面的思路时间复杂度常数更小，Python 实现如下

python 复制代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        if k == 0 or not head or not head.next:
            return head

        _head = head
        total = 1
        while _head.next:
            total += 1
            _head = _head.next
        _head.next = head

        count = total - k % total
        while count:
            _head = _head.next
            count -= 1
        
        res = _head.next
        _head.next = None
        return res

时间复杂度O(n)

空间复杂度O(1)