- 快慢指针+翻转链表
通过快慢指针找到中间节点,然后将后半段进行翻转,然后与前半段进行拼接。
class Solution {
public void reorderList(ListNode head) {
if(head == null || head.next == null) return;
ListNode fast = head.next, slow = head;
while(fast != null && fast.next != null){
slow = slow.next;
fast = fast.next.next;
}
ListNode node = reverse(slow.next);
slow.next = null;
slow = head;
while(slow != null && node != null){
ListNode tmp = node;
node = node.next;
tmp.next = slow.next;
slow.next = tmp;
slow = slow.next.next;
}
}
public ListNode reverse(ListNode head){
if(head == null || head.next == null) return head;
ListNode prev = null;
while(head != null){
ListNode next = head.next;
head.next = prev;
prev = head;
head = next;
}
return prev;
}
}
标签:leetcode143,slow,ListNode,head,fast,next,链表,重排,null
From: https://www.cnblogs.com/xzh-yyds/p/16630430.html