Reverse Linked List 描述 反转一个单链表. 示例: 输入: 1->2->3->4->5->NULL    输出: 5->4->3->2->1->NULL 进阶: 你可以迭代或递归地反转链表.你能否用两种方法解决这道题? 解析 设置三个节点pre.cur.next (1)每次查看cur节点是否为NULL,如果是,则结束循环,获得结果 (2)如果cur节点不是为NULL,则先设置临时变量next为cur的下一个节点 (3)让cur…

Reverse a singly linked list. Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 题意: 如题 思路: 无 代码: /** * Definition for…

反转一个单链表.进阶:链表可以迭代或递归地反转.你能否两个都实现一遍?详见:https://leetcode.com/problems/reverse-linked-list/description/ Java实现: 迭代实现: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ cl…

Reverse a singly linked list. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 反向链表,分别用递归和迭代方式实现. 递归Iteration: 新建一个node(value=任意值, next = None), 用一个变量 next 记录head.next,head.next指向新node.next,新 node.next…

Reverse a singly linked list. Example: Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both? 解法一:(C++)利用迭代的方法依次将链表元素放在新链表的…

题目链接:https://leetcode.com/problems/reverse-linked-list/ 方法一:迭代反转 https://blog.csdn.net/qq_17550379/article/details/80647926讲的很清楚 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), nex…

网址:https://leetcode.com/problems/reverse-linked-list/ 直接参考92:https://www.cnblogs.com/tornado549/p/10639756.html /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; *…

表不支持随机查找,通常是使用next指针进行操作. 206. 反转链表 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ //时间:O(n) 只遍历了一遍链表 //空间:O(1) 开了三个指针的空间 class Solution { public: ListNode*…

206. Reverse Linked List 之前在牛客上的写法: 错误代码: class Solution { public: ListNode* ReverseList(ListNode* pHead) { if(pHead == NULL) return NULL; ListNode* p1 = pHead; ListNode* p2 = pHead->next; ListNode* p3 = pHead->next->next; pHead->next = NULL;…

Question 206. Reverse Linked List Solution 题目大意:对一个链表进行反转 思路: Java实现: public ListNode reverseList(ListNode head) { ListNode newHead = null; while (head != null) { ListNode tmp = head.next; head.next = newHead; newHead = head; head = tmp; } return new…