61. 旋转链表 (opens new window)

English Version (opens new window)

题目描述

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

解法

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。

Python3

# 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 k == 0 or head is None or head.next is None:
            return head
        n, cur = 0, head
        while cur:
            n, cur = n + 1, cur.next
        k %= n
        if k == 0:
            return head
        
        slow = fast = head
        for _ in range(k):
            fast = fast.next
        while fast.next:
            slow, fast = slow.next, fast.next
        
        start = slow.next
        slow.next = None
        fast.next = head
        return start

Java

/**
 * 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 (k == 0 || head == null || head.next == null) {
            return head;
        }
        int n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            ++n;
        }
        k %= n;
        if (k == 0) {
            return head;
        }
        ListNode slow = head, fast = head;
        while (k-- > 0) {
            fast = fast.next;
        }
        while (fast.next != null) {
            slow = slow.next;
            fast = fast.next;
        }

        ListNode start = slow.next;
        slow.next = null;
        fast.next = head;
        return start;
    }
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function rotateRight(head: ListNode | null, k: number): ListNode | null {
    if (k == 0  || head == null || head.next == null) return head;
    // mod n
    let n = 0;
    let p = head;
    while (p != null) {
        ++n;
        p = p.next;
    }
    k %= n;
    if (k == 0) return head;
    
    let fast = head, slow = head;
    for (let i = 0; i < k; ++i) {
        fast = fast.next;
    }
    while (fast.next != null) {
        slow = slow.next;
        fast = fast.next;
    }
    let start = slow.next;
    slow.next = null;
    fast.next = head;
    return start;
};

C++

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (k == 0 || !head || !head->next) {
            return head;
        }
        int n = 0;
        for (ListNode *cur = head; !!cur; cur = cur->next) {
            ++n;
        }
        k %= n;
        if (k == 0) {
            return head;
        }
        ListNode *slow = head, *fast = head;
        while (k-- > 0) {
            fast = fast->next;
        }
        while (fast->next) {
            slow = slow->next;
            fast = fast->next;
        }

        ListNode *start = slow->next;
        slow->next = nullptr;
        fast->next = head;
        return start;
    }
};

C#

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode RotateRight(ListNode head, int k) {
        if (k == 0 || head == null || head.next == null)
        {
            return head;
        }
        var n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next)
        {
            ++n;
        }
        k %= n;
        if (k == 0)
        {
            return head;
        }
        ListNode slow = head, fast = head;
        while (k-- > 0)
        {
            fast = fast.next;
        }
        while (fast.next != null)
        {
            slow = slow.next;
            fast = fast.next;
        }

        ListNode start = slow.next;
        slow.next = null;
        fast.next = head;
        return start;
    }
}

...