[LeetCode] 92. 反转链表 II(Reverse Linked List II)

本文已收录于 LeetCode刷题 系列，共计 28 篇，本篇是第 11 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]

反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。

说明:
1 ≤ m ≤ n ≤ 链表长度。

示例:

输入: 1->2->3->4->5->NULL, m = 2, n = 4
输出: 1->4->3->2->5->NULL

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-linked-list-ii
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″]285179923[/bilibili]

 

[title]简明思路[/title]

[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* reverseBetween(ListNode* head, int m, int n) {
        if (head == nullptr) return head;
        
        ListNode dummy(0);
        ListNode* pre = &dummy;
        pre->next = head;

        for (int i = 1; i < m; i++)
        {
            pre = pre->next;
        }

        ListNode* cur = pre->next;

        for (int i = m; i < n; i++)
        {
            ListNode* future = cur->next;
            cur->next = future->next;
            future->next = pre->next;
            pre->next = future;
        }
        return dummy.next;

    }
};

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* reverseBetween(ListNode* head, int m, int n) {
12. if (head == nullptr) return head;
14. ListNode dummy(0);
15. ListNode* pre = &dummy;
16. pre->next = head;
18. for (int i = 1; i < m; i++)
19. {
20. pre = pre->next;
21. }
23. ListNode* cur = pre->next;
25. for (int i = m; i < n; i++)
26. {
27. ListNode* future = cur->next;
28. cur->next = future->next;
29. future->next = pre->next;
30. pre->next = future;
31. }
32. return dummy.next;
34. }
35. };

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == nullptr) return head;
        
        ListNode dummy(0);
        ListNode* pre = &dummy;
        pre->next = head;

        for (int i = 1; i < m; i++)
        {
            pre = pre->next;
        }

        ListNode* cur = pre->next;

        for (int i = m; i < n; i++)
        {
            ListNode* future = cur->next;
            cur->next = future->next;
            future->next = pre->next;
            pre->next = future;
        }
        return dummy.next;

    }
};

 

作者： 高志远

高志远，24岁，男生 查看高志远的所有文章