1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode current = head;
    ListNode next = null;

    while (current != null) {
        next = current.next; // 保存當前節點的下一個節點
        current.next = prev; // 改變當前節點的指針指向前一個節點
        prev = current; // 將 prev 和 current 向前移動
        current = next;
    }

    head = prev; // 更新頭節點
    return head;
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

def reverseListRecursive(self, head):
    def reverse(cur, prev):
        if cur is None:
            return prev
        else:
            next = cur.next
            cur.next = prev
            return reverse(next, cur)

    return reverse(head, None)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
        next = null;
    }
}

public class Solution {
    public ListNode reverseLinkedList(ListNode head) {
        // 遞歸的基本情況：如果頭節點是 null 或者是最後一個節點，則不需要進一步的反轉
        if (head == null || head.next == null) {
            return head;
        }

        // 遞歸反轉鏈表的剩餘部分
        ListNode reversedListHead = reverseLinkedList(head.next);

        // 將當前節點的下一個節點指向當前節點，從而實現反轉
        head.next.next = head;

        // 斷開當前節點的下一個指針，以避免循環
        head.next = null;

        // 返回反轉後的鏈表頭節點
        return reversedListHead;
    }
}