javascript 解法, 执行用时: 68 ms, 内存消耗: 51.1 MB, 提交时间: 2024-04-20 16:28:11

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} k
 * @return {ListNode}
 */
var rotateRight = function(head, k) {
    if (k === 0 || !head || !head.next) {
        return head;
    }
    let n = 1;
    let cur = head;
    while (cur.next) {
        cur = cur.next;
        n++;
    }

    let add = n - k % n;
    if (add === n) {
        return head;
    }

    cur.next = head;
    while (add) {
        cur = cur.next;
        add--;
    }

    const ret = cur.next;
    cur.next = null;
    return ret;
};

cpp 解法, 执行用时: 12 ms, 内存消耗: 14.9 MB, 提交时间: 2024-04-20 16:27:49

/**
 * 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 == nullptr || head->next == nullptr) {
            return head;
        }
        int n = 1;
        ListNode* iter = head;
        while (iter->next != nullptr) {
            iter = iter->next;
            n++;
        }
        int add = n - k % n;
        if (add == n) {
            return head;
        }
        iter->next = head;
        while (add--) {
            iter = iter->next;
        }
        ListNode* ret = iter->next;
        iter->next = nullptr;
        return ret;
    }
};

java 解法, 执行用时: 0 ms, 内存消耗: 41.4 MB, 提交时间: 2024-04-20 16:27:34

/**
 * 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 = 1;
        ListNode iter = head;
        while (iter.next != null) {
            iter = iter.next;
            n++;
        }
        int add = n - k % n;
        if (add == n) {
            return head;
        }
        iter.next = head;
        while (add-- > 0) {
            iter = iter.next;
        }
        ListNode ret = iter.next;
        iter.next = null;
        return ret;
    }
}

golang 解法, 执行用时: 0 ms, 内存消耗: 2.5 MB, 提交时间: 2024-04-20 16:26:54

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func rotateRight(head *ListNode, k int) *ListNode {
    if k == 0 || head == nil || head.Next == nil {
        return head
    }
    n := 1
    iter := head
    for iter.Next != nil {
        iter = iter.Next
        n++
    }
    add := n - k%n
    if add == n {
        return head
    }
    iter.Next = head
    for add > 0 {
        iter = iter.Next
        add--
    }
    ret := iter.Next
    iter.Next = nil
    return ret
}

python3 解法, 执行用时: 44 ms, 内存消耗: 14.9 MB, 提交时间: 2022-08-02 15:28:14

# 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 not head or not head.next:
            return head
        
        n = 1
        cur = head
        while cur.next:
            cur = cur.next
            n += 1
        
        if (add := n - k % n) == n:
            return head
        
        cur.next = head
        while add:
            cur = cur.next
            add -= 1
        
        ret = cur.next
        cur.next = None
        return ret