https://leetcode.cn/problems/fan-zhuan-lian-biao-lcof/
定义一个函数,输入一个链表的头节点,反转该链表并输出反转后链表的头节点。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
限制:
0 <= 节点个数 <= 5000
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/fan-zhuan-lian-biao-lcof
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1 2 3 4 5 划分子问题 1 2, 2 3, 3 4, 4 5(最后执行,返回5)
递归公式curr->next->next = curr->next;
curr -> next = NULL;
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* reverseList(ListNode* head) { if(head ==NULL || head->next == NULL) return head; ListNode *new_head = reverseList(head->next); head->next->next = head; head->next = NULL; return new_head; } };
标签:24,head,ListNode,Offer,next,链表,curr,NULL From: https://www.cnblogs.com/ygtzds/p/17189830.html