标签：html sub positive memory mod osi int rev 更新
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 the end should remain as it is.
Example:
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
Note:
解题思路：https://leetcode.windliang.cc/leetCode-25-Reverse-Nodes-in-k-Group.html
public ListNode reverseKGroup(ListNode head, int k) { if (head == null) return null; ListNode sub_head = head; ListNode dummy = new ListNode(0); dummy.next = head; ListNode tail = dummy; ListNode toNull = head; while (sub_head != null) { int i = k; //找到子链表的尾部 while (i - 1 > 0) { toNull = toNull.next; if (toNull == null) { return dummy.next; } i--; } ListNode temp = toNull.next; //将子链表断开 toNull.next = null; ListNode new_sub_head = reverse(sub_head); //将倒置后的链表接到 tail 后边 tail.next = new_sub_head; //更新 tail tail = sub_head; //sub_head 由于倒置其实是新链表的尾部 sub_head = temp; toNull = sub_head; //将后边断开的链表接回来 tail.next = sub_head; } return dummy.next; } public ListNode reverse(ListNode head) { ListNode current_head = null; while (head != null) { ListNode next = head.next; head.next = current_head; current_head = head; head = next; } return current_head; }
很好很强大，我很饿
标签：html sub positive memory mod osi int rev 更新
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/10347469.html
