LeetCode 25. Reverse Nodes in k-Group分析代码
LeetCode 25. Reverse Nodes in k-Group分析代码
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. You may not alter the values in the nodes, only nodes itself may be changed. Only constant memory is allowed. For 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 给你一个链表以及一个k,将这个链表从头指针开始每k个翻转一下。 链表元素个数不是k的倍数，最后剩余的不用翻转。 样例 给出链表 1->2->3->4->5 k = 2, 返回 2->1->4->3->5 k = 3, 返回 3->2->1->4->5
分析
我们设计一个翻转k的算法 head -> n1 -> n2 ... nk -> nk+1 => head -> nk -> nk-1 .. n1 -> nk+1 return n1 那么我们只要接着对这个方法传入n1即可，重复调用这个方法，直到遇到null
代码
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
while (true) {
head = reverseK(head, k);
if (head == null) {
break;
}
}
return dummy.next;
}
// head -> n1 -> n2 ... nk -> nk+1
// =>
// head -> nk -> nk-1 .. n1 -> nk+1
// return n1
public ListNode reverseK(ListNode head, int k) {
ListNode nk = head;
for (int i = 0; i < k; i++) {
if (nk == null) {
return null;
}
nk = nk.next;
}
if (nk == null) {
return null;
}
// reverse
ListNode n1 = head.next;
ListNode nkplus = nk.next;
ListNode prev = null;
ListNode curt = n1;
while (curt != nkplus) {
ListNode temp = curt.next;
curt.next = prev;
prev = curt;
curt = temp;
}
// connect
head.next = nk;
n1.next = nkplus;
return n1;
}
}