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# Reverse Nodes in k-Group

ID: 450; hard; K组翻转链表

/**

* Definition for ListNode

* public class ListNode {

* int val;

* ListNode next;

* ListNode(int x) {

* val = x;

* next = null;

* }

* }

*/

public class Solution {

/**

* @param head: a ListNode

* @param k: An integer

* @return: a ListNode

*/

public ListNode reverseKGroup(ListNode head, int k) {

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode cur = dummy;

while (true) {

cur = reverseBetween(cur, k);

if (cur == null) break;

}

return dummy.next;

}

// modified from Reverse Linked List II

public ListNode reverseBetween(ListNode head, int k) {

if (head == null) return head;

// 1. find the k-th node

ListNode kNode = head;

for (int i = 0; i < k; i++) {

kNode = kNode.next;

if (kNode == null) return null;

}

ListNode firstNode = head.next;

ListNode kNextNode = kNode.next;

// 2. reverse

ListNode prev = null, next;

ListNode curr = firstNode;

while (curr != kNextNode) {

next = curr.next;

curr.next = prev;

prev = curr;

curr = next;

}

// 3. connect the lists

head.next = kNode;

firstNode.next = kNextNode;

return firstNode;

}

}

- We use a similar approach used in Reverse Linked List II. Using the helper function
`reverseBetween`

, we reverse`k`

nodes each time and return the end of of the linked list for each`k`

group. If the end node is null, we know that the remaining nodes cannot constitute a`k`

group. - Inside the helper function, note that the
`head`

is always one node before the starting node that is to be reversed. Then, we keep track of the first node that is to be reversed (`firstNode`

) and the node after k-th node (`kNextNode`

) because we need to connect the reversed linked list eventually with a start and an end.

Last modified 2yr ago