[LeetCode] Reverse Linked List 倒置链表】的更多相关文章

Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 之前做到Reverse Linked List II 倒置链表之二的时候我还纳闷怎么只有二没有一呢,原来真是忘了啊,现在才加上,这道题跟之前那道比起来简单不少,题目为了增加些许难度,让我们分别用…
Reverse a linked list. Have you met this question in a real interview? Yes Example For linked list 1->2->3, the reversed linked list is 3->2->1 Challenge Reverse it in-place and in one-pass LeetCode上的原题,请参见我之前的博客Reverse Linked List. 解法一: class…
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…
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 linked list from position m to n. Do it in-place and in one-pass. For example:Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note:Given m, n satisfy the following condition:1 ≤ m ≤ n ≤ lengt…
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…
题目:Reverse Linked List II 题意:Reverse a linked list from position m to n. Do it in-place and in one-pass. 下面这段代码,有两个地方,一个是4.5行的dummy节点设置:另一个是11-14行,局部可视化到全局. ListNode *reverseBetween(ListNode *head, int m, int n) { if(m == n) return head; n -= m; List…
Design your implementation of the linked list. You can choose to use the singly linked list or the doubly linked list. A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointe…
Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 给定一个链表,判断是否有环存在.Follow up: 不使用额外空间. 解法:双指针,一个慢指针每次走1步,一个快指针每次走2步的,如果有环的话,两个指针肯定会相遇. Java: public class Solution { public boolean hasCycle(Li…
翻转一个单链表.这个题目听说很多次了,总感觉肯定不是什么难题. 现在真的有点好高骛远了!总感觉那种很难的算法题才是难题,这种题没必要做.其实眼高手低啊. 这种easy题,我都不能一遍ac,这遇到白板编程也是挂的节奏! 仔细分析,每次翻转一个,要记录被反转的这个的前后节点. 1 -> 2 -> 3 -> 4 用p记录当前要改变其next指针的节点.last 指向前一个节点.pre指向后一个节点. 初始化,last = NULL. struct ListNode* reverse(struc…