/** * Source : https://leetcode.com/problems/reverse-linked-list/ * * * 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? */ public class…

题目:单链表取反 #include <stdlib.h> #include <stdio.h> typedef struct node *list; typedef struct node *position; typedef struct node *ListNode; typedef struct node { int data; position next; }node; static list init_list(void); static void delete_list…

目录 描述 解法一:迭代 思路 Java 实现 Python 实现 复杂度分析 解法二:递归 思路 Java 实现 Python 实现 复杂度分析 更多 LeetCode 题解笔记可以访问我的 github. 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? 解法一:迭代 思路 遍历链表的每个节点,将每个…

题目:2.2.2 Reverse Linked List II 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->nullptr, m = 2 and n = 4, return 1->4->3->2->5->nullptr. Note: Given m, n satisfy t…

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…

https://leetcode.com/problems/reverse-linked-list/ 思路很简单,分别设置三个结点,之后依次调整结点1和结点2的指向关系. Before: pre -> nxt -> nxtnxt -> .....  Here current = pre,nxt = pre->next, nxtnxt = nxt->next. After:   pre <- nxt     nxtnxt -> .....  Here current…

Leetcode Solutions Language: javascript c mysql Last updated: 2019-01-04 https://github.com/nusr/leetcode # Problems Solutions Difficulty Acceptance Paid-Only 001 two-sum c,javascript Easy 39.69% No 002 add-two-numbers javascript Medium 30.01% No 007…

面试题16:反转链表 提交网址: http://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=13&tqId=11168 或 https://leetcode.com/problems/reverse-linked-list/ Total Accepted: 101523  Total Submissions: 258623  Difficulty: Easy Reverse a singly linked lis…

一.通用方法以及题目分类 0.遍历链表 方法代码如下,head可以为空: ListNode* p = head; while(p!=NULL) p = p->next; 可以在这个代码上进行修改,比如要计算链表的长度: ListNode* p = head; ; while(p!=NULL){ num++; p = p->next; } 如果要找到最后的节点,可以更改while循环中的条件,只不过需要加上head为NULL时的判断 if(!head) return head; ListNode…

最新更新时间 11:22:29 8. String to Integer (atoi) public static int myAtoi(String str) { // 1字符串非空判断 ""||" " if (str.isEmpty() || str.trim().isEmpty()) { return 0; } int index = 0; int sign = 1; int total = 0; //1检测第一个非空字符串是什么 while (str.cha…