• Reverse Linked List
  • Solution (Traverse the linked list, pointing the next of each node to the previous node)
  • Step
  • Code
• Reverse Linked List II
  • Solution (Iterate and insert the following node of curr to the position after pre)
  • Step
  • Code

Reverse Linked List

Solution (Traverse the linked list, pointing the next of each node to the previous node)

1. Save the next of the current node, next = curr.next
2. Point the next of the current node to the previous node, curr.next = prev
3. Move the forward pointer back to prev = curr
4. Move the current pointer back to curr = next

Step

1.Initial list and pre node

2.Initial list and next node, next = curr.next

3.The current.next points to the pre node, curr.next = prev

4.The pre node points to the current node, prev = curr

5.The current node points to the next node, curr = next

6.The next node moves to the current.next, next = curr.next

7.Loop traversal

Code

package LinkedLists;

/**
 * @author zhengstars
 * @date 2023/08/19
 */
public class ReverseLinkedList {
    public static class ListNode {
        int value;
        ListNode next;

        public ListNode(int value) {
            this.value = value;
            this.next = null;
        }
    }

    public static ListNode reverseLinkedList(ListNode head) {
        ListNode prev = null;

        ListNode curr = head;

        ListNode next = null;

        while (curr != null) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;

        }
        return prev;
    }
}

Reverse Linked List II

Solution (Iterate and insert the following node of curr to the position after pre)

1. Save the next node of the current node, ListNode next = curr.next
2. Let the next pointer of the current node point to the node pointed by next’s next pointer. curr.next = next.next;
3. Point the next of the next node to the node pointed by pre’s next pointer . next.next = pre.next
4. Move the next pointer to the position after the pre pointer, by setting the next pointer of the pre poniter to the next pointer. pre.next = next;

Step

1.Initial List and Pre(head) Node

left = 2, right = 4. After the pre loop, pre points to 1 and curr points to 2.

2.Initial List and Next Node, next = curr.next

Before: curr.next = 3

3.The current.next points to the pre.next node, curr.next = next.next

After: curr.next = 4

4.The next.next node points to the pre.next. next.next = pre.next

next.next = pre.next; curr.next = 4 next.next = 2 pre.next = 2

5.The pre node points to the next node, pre.next = next

curr.next = 4 next.next = 2 pre.next = 3

6.Loop1

7.Repeat step 2 to 5 , loop traversal, until to meet the end of range

ListNode next = 4 curr.next = 4 next.next = 5 pre.next = 3

8.Loop traversal

Code

package LinkedLists;

/**
 * @author zhengstars
 * @date 2025/09/21
 */
public class ReverseLinkedListII {

    public static class ListNode {
        int value;
        ListNode next;

        public ListNode(int value) {
            this.value = value;
            this.next = null;
        }
    }

    public static ListNode reverseBetween(ListNode head, int left, int right) {
        if (head == null || left == right) {
            return head;
        }

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;

        for (int i = 1; i < left; i++) {
            pre = pre.next;
        }

        ListNode curr = pre.next;

        for (int i = 0; i < right - left; i++) {
            ListNode next = curr.next;
            curr.next = next.next;
            next.next = pre.next;
            pre.next = next;
        }

        return dummy.next;
    }
}