【12_206】Reverse Linked List

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【12_206】Reverse Linked List

本来没想出来,刚才突然想到,可以用“头插法”来反转 Reverse Linked List My Submissions Question Total Accepted: 66556 Total Submissions: 182891 Difficulty: Easy Reverse a singly linked list. click to show more hints. Discuss中的递归解法: public ListNode reverseList(ListNode head) {…

【leetcode】Reverse Linked List II

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->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note:Given m, n satisfy the following cond…

【leetcode】Reverse Linked List II （middle）

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. 思路: 好困啊,脑子晕晕的. 转了半天AC了.但写的很罗嗦,要学习大神的写法. 注意翻转的写法. 用伪头部大神14行简洁代码 L…

【leetcode】Reverse Linked List（easy）

Reverse a singly linked list. 思路:没啥好说的.秒... ListNode* reverseList(ListNode* head) { ListNode * rList = NULL, * tmp = NULL; while(head != NULL) { tmp = rList; rList = head; head = head->next; rList->next = tmp; } return rList; }…

【链表】 Reverse Linked List II

题目: 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 ≤ l…

【Leetcode】【Easy】Reverse Linked List

题目: Reverse a singly linked list. 解题: 反转单链表,不再多介绍了. 如果会“先条件->定参数->确定不变式->验证后条件”的思维方法,一定会bug free. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class So…

【LeetCode92】Reverse Linked List II★★

题目描述: 解题思路: 题目大意:给定一个链表,反转第m到第n个结点部分,m.n满足1 ≤ m ≤ n ≤ length of list. 解题思路参照LeetCode206题,用迭代法,不过要注意以下几点: (a):为方便操作,新建一个辅助结点dummy,使其下一个结点指向头节点. (b):维护4个指针:pre.current.then.then_next,pre指向第m个结点的前一个结点,current指向当前操作的位于区间[m,n]的结点,then指向current的下一个结点,then_…