Reorder List

描述

Given a singly linked list L:L0→L1→⋯→Ln−1→LnL: L0 \rightarrow L_1 \rightarrow \cdots \rightarrow L{n-1} \rightarrow Ln,reorder it to: L0→Ln→L1→Ln−1→L2→Ln−2→⋯L_0 \rightarrow L_n \rightarrow L_1 \rightarrow L{n-1} \rightarrow L2 \rightarrow L{n-2} \rightarrow \cdots

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}.

分析

题目规定要in-place，也就是说只能使用O(1)的空间。

可以找到中间节点，断开，把后半截单链表reverse一下，再合并两个单链表。

代码

  1. // Reorder List
  2. // 时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4.     public void reorderList(ListNode head) {
  5.         if (head == null || head.next == null) return;
  7.         ListNode slow = head, fast = head, prev = null;
  8.         while (fast != null && fast.next != null) {
  9.             prev = slow;
  10.             slow = slow.next;
  11.             fast = fast.next.next;
  12.         }
  13.         prev.next = null; // cut at middle
  15.         slow = reverse(slow);
  17.         // merge two lists
  18.         ListNode curr = head;
  19.         while (curr.next != null) {
  20.             ListNode tmp = curr.next;
  21.             curr.next = slow;
  22.             slow = slow.next;
  23.             curr.next.next = tmp;
  24.             curr = tmp;
  25.         }
  26.         curr.next = slow;
  27.     }
  29.     ListNode reverse(ListNode head) {
  30.         if (head == null || head.next == null) return head;
  32.         ListNode prev = head;
  33.         for (ListNode curr = head.next, next = curr.next; curr != null;
  34.             prev = curr, curr = next, next = next != null ? next.next : null) {
  35.                 curr.next = prev;
  36.         }
  37.         head.next = null;
  38.         return prev;
  39.     }
  40. }

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/linear-list/linked-list/reorder-list.html