题目: Reverse a linked list. Example For linked list 1->2->3, the reversed linked list is 3->2->1 题解: Solution 1 () class Solution { public: ListNode *reverse(ListNode *head) { if (!head) { return head; } ListNode *prev = nullptr; while (head) {…

题目: Reverse a linked list from position m to n. Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length of list. Example Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. 题解: Solution 1 () class S…

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…

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? 解法一:非递归 /** * Definition for singly-linked list. * struct ListNode { * int va…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 迭代 递归 日期 题目地址:https://leetcode.com/problems/reverse-linked-list-ii/description/ 题目描述 Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of…

题目: Reverse a singly linked list. 提示: 此题不难,可以用迭代或者递归两种方法求解.记得要把原来的链表头的next置为NULL: 代码: 迭代: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { pub…

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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 迭代 递归 日期 [LeetCode] 题目地址:https://leetcode.com/problems/reverse-linked-list/ Total Accepted: 105474 Total Submissions: 267077 Difficulty: Easy 题目描述 Reverse a singly linked list.…

链表反转,一发成功~ /** * 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) { //就地反转 if(head == NULL || head ->n…

The task is reversing a list in range m to n(92) or a whole list(206). All in one : U need three pointers to achieve this goal. 1) Pointer to last value 2) Pointer to cur p value 3) Pointer to next value Here, showing my code wishes can help u. Of co…