题解 | #反转链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head ListNode类 
     * @return ListNode类
     */
    ListNode* ReverseList(ListNode* head) {
        // write code here
    #if 0
        //迭代反转
        if(head == nullptr || head->next == nullptr)
            return head;
        
        ListNode* beg = nullptr;
        ListNode* mid = head;
        ListNode* end = head->next;
        while (1) {
            
            //修改mid所指节点指向beg
            mid->next = beg;

            //判断end
            if(end == nullptr)
                break;

            //调整向后移动
            beg = mid;
            mid = end;
            end = end->next;
        }
        //修改head头指针的指向
        head = mid;
        return head;
    #else
        //接地逆置
        if(head == nullptr || head->next == nullptr)
            return head;
        
        ListNode* prev = nullptr;
        ListNode* curr = head;
        while (curr) {
            ListNode* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    #endif
    }
};