LeetCode 使用原地算法实现单链表的特殊旋转
问题描述:
将给定的单链表L: L 0→L 1→…→L n-1→L n,
重新排序为: L 0→L n →L 1→L n-1→L 2→L n-2→…
要求使用原地算法,并且不改变节点的值
例如:
对于给定的单链表{1,2,3,4},将其重新排序为{1,4,2,3}.
Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…
You must do this in-place without altering the nodes' values.
For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.
解题思路:考虑把原始链表拆分为两个,利用快慢指针可以找到链表的中间节点
再把后面部分的链表实现翻转,使用原地算法
再把两个链表合并为一个链表即实现上述要求
/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public void reorderList(ListNode head) { if(head == null || head.next == null){ return; } ListNode slow = head; ListNode fast = head; while(fast.next != null && fast.next.next != null){ slow = slow.next; fast = fast.next.next; } ListNode mid = slow.next; slow.next = null; ListNode after = getReverse(mid); getAdd(head, after); } public ListNode getReverse(ListNode node){ ListNode pre = null; while(node != null){ ListNode temp = node.next; node.next = pre; pre = node; node = temp; } return pre; } public void getAdd(ListNode node1,ListNode node2){ while(node1 != null && node2 !=null){ ListNode f = node1.next; ListNode a = node2.next; node1.next = node2; node1 = f; node2.next = node1; node2 = a; } } }