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Reorder List

描述

Given a singly linked list L:L0L1Ln1LnL: L_0 \rightarrow L_1 \rightarrow \cdots \rightarrow L_{n-1} \rightarrow L_n, reorder it to: L0LnL1Ln1L2Ln2L_0 \rightarrow L_n \rightarrow L_1 \rightarrow L_{n-1} \rightarrow L_2 \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

// Reorder List
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null) return;

ListNode slow = head, fast = head, prev = null;
while (fast != null && fast.next != null) {
prev = slow;
slow = slow.next;
fast = fast.next.next;
}
prev.next = null; // cut at middle

slow = reverse(slow);

// merge two lists
ListNode curr = head;
while (curr.next != null) {
ListNode tmp = curr.next;
curr.next = slow;
slow = slow.next;
curr.next.next = tmp;
curr = tmp;
}
curr.next = slow;
}

ListNode reverse(ListNode head) {
if (head == null || head.next == null) return head;

ListNode prev = head;
for (ListNode curr = head.next, next = curr.next; curr != null;
prev = curr, curr = next, next = next != null ? next.next : null) {
curr.next = prev;
}
head.next = null;
return prev;
}
}