给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。
示例 1:
输入: 1->2->3->4->5->NULL, k = 2 输出: 4->5->1->2->3->NULL 解释: 向右旋转 1 步: 5->1->2->3->4->NULL 向右旋转 2 步: 4->5->1->2->3->NULL
示例 2:
输入: 0->1->2->NULL, k = 4 输出:2->0->1->NULL
解释: 向右旋转 1 步: 2->0->1->NULL 向右旋转 2 步: 1->2->0->NULL 向右旋转 3 步:0->1->2->NULL
向右旋转 4 步:2->0->1->NULL
将链表右半部分的 k 的节点拼接到 head 即可。
注:k 对链表长度 n 取余,即 k %= n
。
# 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: ListNode, k: int) -> ListNode:
if head is None or head.next is None or k == 0:
return head
n = 0
cur = head
while cur:
n += 1
cur = cur.next
k %= n
if k == 0:
return head
p = q = head
for i in range(k):
q = q.next
while q.next:
p, q = p.next, q.next
start = p.next
p.next = None
q.next = head
return start
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null || head.next == null) {
return head;
}
int n = 0;
ListNode cur = head;
while (cur != null) {
++n;
cur = cur.next;
}
k %= n;
if (k == 0) {
return head;
}
ListNode p = head, q = head;
for (int i = 0; i < k; ++i) {
q = q.next;
}
while (q.next != null) {
p = p.next;
q = q.next;
}
ListNode start = p.next;
p.next = null;
q.next = head;
return start;
}
}