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61. 旋转链表

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

方法一：快慢指针 + 链表拼接

我们先判断链表节点数是否小于 \(2\)，如果是，直接返回 \(head\) 即可。

否则，我们先统计链表节点数 \(n\)，然后将 \(k\) 对 \(n\) 取模，得到 \(k\) 的有效值。

如果 \(k\) 的有效值为 \(0\)，说明链表不需要旋转，直接返回 \(head\) 即可。

否则，我们用快慢指针，让快指针先走 \(k\) 步，然后快慢指针同时走，直到快指针走到链表尾部，此时慢指针的下一个节点就是新的链表头节点。

最后，我们将链表拼接起来即可。

时间复杂度 \(O(n)\)，其中 \(n\) 是链表节点数，空间复杂度 \(O(1)\)。

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

        ans = slow.next
        slow.next = None
        fast.next = head
        return ans

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

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

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/**
 * 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
    }
    cur := head
    n := 0
    for cur != nil {
        cur = cur.Next
        n++
    }
    k %= n
    if k == 0 {
        return head
    }
    fast, slow := head, head
    for i := 0; i < k; i++ {
        fast = fast.Next
    }
    for fast.Next != nil {
        fast = fast.Next
        slow = slow.Next
    }
    ans := slow.Next
    slow.Next = nil
    fast.Next = head
    return ans
}

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/**
 * 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 (!head || !head.next) {
        return head;
    }
    let cur = head;
    let n = 0;
    while (cur) {
        cur = cur.next;
        ++n;
    }
    k %= n;
    if (k === 0) {
        return head;
    }
    let fast = head;
    let slow = head;
    while (k--) {
        fast = fast.next;
    }
    while (fast.next) {
        fast = fast.next;
        slow = slow.next;
    }
    const ans = slow.next;
    slow.next = null;
    fast.next = head;
    return ans;
}

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// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn rotate_right(mut head: Option<Box<ListNode>>, mut k: i32) -> Option<Box<ListNode>> {
        if head.is_none() || k == 0 {
            return head;
        }
        let n = {
            let mut cur = &head;
            let mut res = 0;
            while cur.is_some() {
                cur = &cur.as_ref().unwrap().next;
                res += 1;
            }
            res
        };
        k = k % n;
        if k == 0 {
            return head;
        }

        let mut cur = &mut head;
        for _ in 0..n - k - 1 {
            cur = &mut cur.as_mut().unwrap().next;
        }
        let mut res = cur.as_mut().unwrap().next.take();
        cur = &mut res;
        while cur.is_some() {
            cur = &mut cur.as_mut().unwrap().next;
        }
        *cur = head.take();
        res
    }
}

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

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