[LeetCode] 61. 旋转链表(Rotate List)

本文已收录于 LeetCode刷题 系列，共计 28 篇，本篇是第 6 篇

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• 42. 接雨水 (Trapping Rain Water)
• [LeetCode] 2. 两数相加 2. Add Two Numbers
• [LeetCode] 19. 删除链表的倒数第N个节点 19. Remove Nth Node From End of List
• [LeetCode] 21. 合并两个有序链表(Merge Two Sorted Lists)
• [LeetCode] 24. 两两交换链表中的节点(Swap Nodes in Pairs)
• [LeetCode] 61. 旋转链表(Rotate List)
• [LeetCode] 82. 删除排序链表中的重复元素 II(Remove Duplicates from Sorted List II)
• [LeetCode] 83. 删除排序链表中的重复元素(Remove Duplicates from Sorted List)
• [LeetCode] 86. 分隔链表(Partition List)
• [LeetCode] 206. 反转链表(Reverse Linked List)
• [LeetCode] 92. 反转链表 II(Reverse Linked List II)
• [LeetCode] 147. 对链表进行插入排序(Insertion Sort List)
• [LeetCode] 203. 移除链表元素(Remove Linked List Elements)
• [LeetCode] 160. 相交链表(Intersection of Two Linked Lists)
• [LeetCode] 237. 删除链表中的节点 (Delete Node in a Linked List)
• [LeetCode] 328. 奇偶链表 (Odd Even Linked List)
• [LeetCode] 876. 链表的中间结点 (Middle of the Linked List)
• [LeetCode] 1290. 二进制链表转整数 (Convert Binary Number in a Linked List to Integer)
• [LeetCode] 141. 环形链表 (Linked List Cycle)
• [LeetCode] 142. 环形链表 II (Linked List Cycle II)
• [LeetCode] 283. 移动零 (Move Zeroes)
• [LeetCode] 496. 下一个更大元素 I (Next Greater Element I)
• [LeetCode] 503. 下一个更大元素 II (Next Greater Element II)
• [LeetCode] 456. 132模式 (132 Pattern)
• [LeetCode] 20. 有效的括号 (Valid Parentheses)
• [LeetCode] 94. 二叉树的中序遍历 (Binary Tree Inorder Traversal)
• [LeetCode] 144. 二叉树的前序遍历 (Binary Tree Preorder Traversal)
• [LeetCode] 145. 二叉树的后序遍历 (Binary Tree Postorder Traversal)

[title]题目[/title]

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/rotate-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

[title]视频讲解[/title]

[bilibili cid=”” page=”1″]752530187[/bilibili]

 

[title]代码[/title]

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (head == nullptr)
        {
            return head;
        }

        ListNode* fast = head;
        ListNode* slow = head;

        //求链表长度
        int n = 1;
        ListNode* len = head;
        while (len->next != nullptr)
        {
            n++;
            len = len->next;
        }

        k = k % n;
        if (k == 0)
        {
            return head;
        }

        for (int i = 0; i < k; i++)
        {
            fast = fast->next;
        }
        while (fast->next != nullptr)
        {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode* newHead = slow->next;
        slow->next = nullptr;
        fast->next = head;

        return newHead;
    }
};

1. /**
2. * Definition for singly-linked list.
3. * struct ListNode {
4. * int val;
5. * ListNode *next;
6. * ListNode(int x) : val(x), next(NULL) {}
7. * };
8. */
9. class Solution {
10. public:
11. ListNode* rotateRight(ListNode* head, int k) {
12. if (head == nullptr)
13. {
14. return head;
15. }
17. ListNode* fast = head;
18. ListNode* slow = head;
20. //求链表长度
21. int n = 1;
22. ListNode* len = head;
23. while (len->next != nullptr)
24. {
25. n++;
26. len = len->next;
27. }
29. k = k % n;
30. if (k == 0)
31. {
32. return head;
33. }
35. for (int i = 0; i < k; i++)
36. {
37. fast = fast->next;
38. }
39. while (fast->next != nullptr)
40. {
41. fast = fast->next;
42. slow = slow->next;
43. }
44. ListNode* newHead = slow->next;
45. slow->next = nullptr;
46. fast->next = head;
48. return newHead;
49. }
50. };

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (head == nullptr)
        {
            return head;
        }

        ListNode* fast = head;
        ListNode* slow = head;

        //求链表长度
        int n = 1;
        ListNode* len = head;
        while (len->next != nullptr)
        {
            n++;
            len = len->next;
        }

        k = k % n;
        if (k == 0)
        {
            return head;
        }

        for (int i = 0; i < k; i++)
        {
            fast = fast->next;
        }
        while (fast->next != nullptr)
        {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode* newHead = slow->next;
        slow->next = nullptr;
        fast->next = head;

        return newHead;
    }
};

 

作者： 高志远

高志远，24岁，男生 查看高志远的所有文章