【Leetcode】143. Reorder List

Question：

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

Tips：

给定一个单链表，将链表重新排序，注意不能改变结点的值。

排序规则如下：

L₀→L₁→…→L_n-1→L_n,
L₀→L_n→L₁→L_n-1→L₂→L_n-2→…

思路：

重新排序后的链表，前1/2 结点相对顺序不变，而后半部分是逆序。所以我的思路是先将后半部分结点翻转，变为逆序，再将后半部分结点依次插入到前半部分中去。

大致分为三部分:

（1）找到链表的中间位置，将链表分为两部分。

（2）将第二部分链表逆序

（3）将第二部分所有节点依次插入到前半部分结点之间。

代码：

public void reorderList(ListNode head) {if (head == null || head.next == null)return;// Find the part2;第二部分是从slow.next开始的ListNode slow = head;ListNode fast = head;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}System.out.println("slow"+slow.val);ListNode mid = slow.next;slow.next = null;System.out.println("mid"+mid.val);// 将第二部分翻转;ListNode pre = null;ListNode cur = mid;while (cur != null) {if (cur.next != null) {ListNode next = cur.next;System.out.println("next"+next.val);cur.next = pre;pre = cur;cur = next;} else {cur.next = pre;pre = cur;cur=null;}}System.out.println("pre"+pre.val);// append one by one;ListNode p1 = head;ListNode p2 = pre;while (p2 != null) {ListNode n1 = p1.next;ListNode n2 = p2.next;p1.next = p2;p2.next = n1;p1 = p1.next.next;p1 = n1;p2 = n2;}    //printwhile (head != null) {System.out.println(head.val);head = head.next;}}

代码中的一些输出是为了验证结果的正确性提交时可删除。leetcode提交版版代码如下：

public void reorderList(ListNode head) {if (head == null || head.next == null)return;// Find the part2;第二部分是从slow.next开始的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;// 将第二部分翻转;ListNode pre = null;ListNode cur = mid;while (cur != null) {if (cur.next != null) {ListNode next = cur.next;cur.next = pre;pre = cur;cur = next;} else {cur.next = pre;pre = cur;cur=null;}}// append one by one;ListNode p1 = head;ListNode p2 = pre;while (p2 != null) {ListNode n1 = p1.next;ListNode n2 = p2.next;p1.next = p2;p2.next = n1;p1 = p1.next.next;p1 = n1;p2 = n2;}}

转载于:https://www.cnblogs.com/yumiaomiao/p/8479734.html

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