題目描述
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.
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
解題思路
每k個(gè)一組翻轉(zhuǎn)列表搂漠，核心在于reverse method， 具體看圖解
Java代碼實(shí)現(xiàn)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || k == 1) {
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode pre = dummy;
int looper = 0;
ListNode traverse = head;
while (traverse != null) {
looper ++;
if (looper % k == 0) {
pre = reverse(pre, traverse.next);
traverse = pre.next;
} else {
traverse = traverse.next;
}
}
return dummy.next;
}
/*
* 0->1->2->3->4->5->6
* | |
* pre next
*
* after calling pre = reverse(pre, next)
*
* 0->3->2->1->4->5->6
* | |
* pre next
*/
public ListNode reverse(ListNode pre, ListNode next) {
ListNode last = pre.next;
ListNode curr = last.next;
while (curr != next) {
last.next = curr.next;
curr.next = pre.next;
pre.next = curr;
curr = last.next;
}
return last;
}
}
