题解 | 反转链表

class Solution {
public:
    ListNode* ReverseList(ListNode* pHead) {
        // 如果链表为空或只有一个节点，直接返回
        if (!pHead || !pHead->next) {
            return pHead;
        }
        
        ListNode* prev = nullptr;
        ListNode* curr = pHead;
        ListNode* next = nullptr;
        
        while (curr) {
            // 保存下一个节点
            next = curr->next;
            // 反转当前节点的指针
            curr->next = prev;
            // 移动prev和curr指针
            prev = curr;
            curr = next;
        }
        
        return prev;
    }
};

public class Solution {
    public ListNode ReverseList(ListNode head) {
        // 如果链表为空或只有一个节点，直接返回
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode prev = null;
        ListNode curr = head;
        ListNode next = null;
        
        while (curr != null) {
            // 保存下一个节点
            next = curr.next;
            // 反转当前节点的指针
            curr.next = prev;
            // 移动prev和curr指针
            prev = curr;
            curr = next;
        }
        
        return prev;
    }
}

class Solution:
    def ReverseList(self, head):
        # 如果链表为空或只有一个节点，直接返回
        if not head or not head.next:
            return head
            
        prev = None
        curr = head
        
        while curr:
            # 保存下一个节点
            next = curr.next
            # 反转当前节点的指针
            curr.next = prev
            # 移动prev和curr指针
            prev = curr
            curr = next
            
        return prev