Java for LeetCode 143 Reorder List

標籤：

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}.

解題思路一：

每次將Ln換到前面，得到L0→Ln→L1→L2→L3→，然後對L1使用相同操作，JAVA實現如下：

public void reorderList(ListNode head) {ListNode headCopy = head;while (headCopy != null && headCopy.next != null&& headCopy.next.next != null) {ListNode temp = headCopy;while (temp.next.next != null)temp = temp.next;temp.next.next = headCopy.next;headCopy.next = temp.next;temp.next = null;temp = headCopy.next.next;headCopy=headCopy.next.next;}}

 結果TLE

解題思路二：

空間換時間，將所有的node存到一個list中，然後每次操作list頭尾兩個node即可，JAVA實現如下：

    public void reorderList(ListNode head) {LinkedList<ListNode> list = new LinkedList<ListNode>();ListNode headCopy = head,end = head;while (headCopy != null) {list.add(headCopy);headCopy = headCopy.next;}while(list.size()>2){headCopy=list.poll();end=list.get(list.size()-1);list.remove(list.size()-1);headCopy.next=end;end.next=list.peek();list.get(list.size()-1).next=null;}    }

 

Java for LeetCode 143 Reorder List