0061. 旋转链表
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
- 求单链表的倒数第 N 个节点
- 1.采用快慢指
- 2.快指针与慢指针都以每步一个节点的速度向后遍历
- 3.快指针比慢指针先走 N 步
- 4.当快指针到达终点时，慢指针正好是倒数第 N 个节点
- 伪代码
快指针 = head;
慢指针 = head;
while (快指针.next) {
if (N-- <= 0) {
慢指针 = 慢指针.next;
}
快指针 = 快指针.next;
}
- 语言支持: JS
JS Code:
let slow = (fast = head);
while (fast.next) {
if (k-- <= 0) {
slow = slow.next;
}
fast = fast.next;
}
- 1.获取单链表的倒数第 N + 1 与倒数第 N 个节点
- 2.将倒数第 N + 1 个节点的 next 指向 null
- 3.将链表尾节点的 next 指向 head
- 4.返回倒数第 N 个节点
例如链表 A -> B -> C -> D -> E 右移 2 位，依照上述步骤为：
- 1.获取节点 C 与 D
- 2.A -> B -> C -> null, D -> E
- 3.D -> E -> A -> B -> C -> nul
- 4.返回节点 D
注意：假如链表节点长度为 len， 则右移 K 位与右移动 k % len 的效果是一样的 就像是长度为 1000 米的环形跑道， 你跑 1100 米与跑 100 米到达的是同一个地点
- 伪代码
获取链表的长度
k = k % 链表的长度
获取倒数第k + 1,倒数第K个节点与链表尾节点
倒数第k + 1个节点.next = null
链表尾节点.next = head
return 倒数第k个节点
- 语言支持: JS, JAVA, Python, CPP, Go, PHP
JS Code:
var rotateRight = function (head, k) {
if (!head || !head.next) return head;
let count = 0,
now = head;
while (now) {
now = now.next;
count++;
}
k = k % count;
let slow = (fast = head);
while (fast.next) {
if (k-- <= 0) {
slow = slow.next;
}
fast = fast.next;
}
fast.next = head;
let res = slow.next;
slow.next = null;
return res;
};
JAVA Code:
class Solution {
public ListNode rotateRight(ListNode head, int k) {
if(head == null || head.next == null) return head;
int count = 0;
ListNode now = head;
while(now != null){
now = now.next;
count++;
}
k = k % count;
ListNode slow = head, fast = head;
while(fast.next != null){
if(k-- <= 0){
slow = slow.next;
}
fast = fast.next;
}
fast.next = head;
ListNode res = slow.next;
slow.next = null;
return res;
}
}
Python Code:
class Solution:
def rotateRight(self, head: ListNode, k: int) -> ListNode:
# 双指针
if head:
p1 = head
p2 = head
count = 1
i = 0
while i < k:
if p2.next:
count += 1
p2 = p2.next
else:
k = k % count
i = -1
p2 = head
i += 1
while p2.next:
p1 = p1.next
p2 = p2.next
if p1.next:
tmp = p1.next
else:
return head
p1.next = None
p2.next = head
return tmp
CPP Code:
class Solution {
int getLength(ListNode *head) {
int len = 0;
for (; head; head = head->next, ++len);
return len;
}
public:
ListNode* rotateRight(ListNode* head, int k) {
if (!head) return NULL;
int len = getLength(head);
k %= len;
if (k == 0) return head;
auto p = head, q = head;
while (k--) q = q->next;
while (q->next) {
p = p->next;
q = q->next;
}
auto h = p->next;
q->next = head;
p->next = NULL;
return h;
}
};
Go Code:
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func rotateRight(head *ListNode, k int) *ListNode {
if head == nil || head.Next == nil {
return head
}
n := 0
p := head
for p != nil {
n++
p = p.Next
}
k = k % n
// p 为快指针, q 为慢指针
p = head
q := head
for p.Next!=nil {
p = p.Next
if k>0 {
k--
} else {
q = q.Next
}
}
// 更新指针
p.Next = head
head = q.Next
q.Next = nil
return head
}
PHP Code:
/**
* Definition for a singly-linked list.
* class ListNode {
* public $val = 0;
* public $next = null;
* function __construct($val) { $this->val = $val; }
* }
*/
class Solution
{
/**
* @param ListNode $head
* @param Integer $k
* @return ListNode
*/
function rotateRight($head, $k)
{
if (!$head || !$head->next) return $head;
$p = $head;
$n = 0;
while ($p) {
$n++;
$p = $p->next;
}
$k = $k % $n;
$p = $q = $head; // $p 快指针; $q 慢指针
while ($p->next) {
$p = $p->next;
if ($k > 0) $k--;
else $q = $q->next;
}
$p->next = $head;
$head = $q->next;
$q->next = null;
return $head;
}
}
复杂度分析
- 时间复杂度：节点最多只遍历两遍，时间复杂度为$O(N)$
- 空间复杂度：未使用额外的空间，空间复杂度$O(1)$
最近更新 2yr ago