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.

【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.】的更多相关文章

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.

我用String代替了链表显示,本题的大意是每k个进行逆序处理,剩下的不够k个的就按照原顺序保留下来. public class ReverseNodes { public static void main(String[] args) { String str = "1->2->3->4->5->6->7->8->9->10->11->12->13->14->15"; String[] strArra…

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

class Solution { public: ListNode *reverseKGroup(ListNode *head, int k) { if (!head || !(head->next) || k < 2) return head; // count k nodes ListNode *nextgp = head; for (int i = 0; i < k; i++) if (nextgp) nextgp = nextgp->next; else return he…

[Linked List]Reverse Nodes in k-Group

Total Accepted: 48614 Total Submissions: 185356 Difficulty: Hard 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 rem…

【LeetCode】9 & 234 & 206 - Palindrome Number & Palindrome Linked List & Reverse Linked List

9 - Palindrome Number 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…

cc150 Chapter 2 | Linked Lists 2.6 Given a circular linked list, implement an algorithm which returns node at the beginning of the loop.

2.6Given a circular linked list, implement an algorithm which returns the node at the beginning of the loop. 快指针和慢指针一起在头指针开始移动,快指针每次移动两步,慢指针每次移动一步,直到相遇, 相遇后,把慢指针指向头指针,两个指针一起往后移动一步.直到相遇,相遇的地方就是循环的开始(如果相遇,说明是有循环的节点,这时快指针走的路程是慢指针的两倍,快:开始的k距离,和一圈(或者n圈)循…