题解 | 反转链表

class Solution {
public:
    ListNode* ReverseList(ListNode* head) {
        ListNode* prev = nullptr;
        ListNode* current = head;
        
        while(current) {
            ListNode* nextTemp = current->next; // 保存下一个节点
            current->next = prev;               // 反转指针
            prev = current;                     // 移动prev
            current = nextTemp;                 // 移动current
        }
        
        return prev; // 当current为null时，prev就是新头节点
    }
};