题目: Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only…

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nod…

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nod…

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in…

Reverse Nodes in k-Group Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the valu…

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nod…

本题是本人字节跳动一面考的算法题原题是有序数组,一时没想到怎么解决数组的问题,但是如果给的是有序链表数组,则可以用下面的方法解决 可以利用最小堆完成,时间复杂度是O(nklogk),具体过程如下: 创建一个大小为k的最小堆,堆中元素为k个链表中的每个链表的第一个元素重复下列步骤每次从堆中取出最小元素(堆顶元素),并将其存入输出数组中用堆顶元素所在链表元素的下一元素将堆顶元素替换掉,初始化最小堆的时间复杂度O(k),总共有kn次循环,每次循环调整最小堆的时间复杂度是O(logk),所以总的时间复杂…

题意:给一个单链表,每k个节点就将这k个节点反置,若节点数不是k的倍数,则后面不够k个的这一小段链表不必反置. 思路:递归法.每次递归就将k个节点反置,将k个之后的链表头递归下去解决.利用原来的函数接口即可,不用重新定义. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; *…

# Definition for singly-linked list. # class ListNode(object): # def __init__(self, x): # self.val = x # self.next = None class Solution(object): def reverseKGroup(self, head, k): """ :type head: ListNode :type k: int :rtype: ListNode "…

题目: Determine whether an integer is a palindrome. Do this without extra space. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra space. You could…