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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 reverseknodes each time and return the end of of the linked list for eachkgroup. If the end node is null, we know that the remaining nodes cannot constitute akgroup. - Inside the helper function, note that the
headis 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