标签：des style blog color io for
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}.
题解：看起来最后变换得到的链表顺序很奇怪，起始很简单，就是把链表从中间折断，后半部分反转后和前半部分做归并即可。不过这个归并不是归并排序每次从两个链表中取较小的值，而是两个链表相互交错即可。
例如1->2->3->4，从中间折断得到两个链表1->2和3->4，后面一个链表反转得到4->3，然后和第一个链表交错归并得到1->3->2->4;
再例如1->2->3->4->5，从中间折断得到两个链表1->2->3和4->5，后一个链表反转得到5->4，和第一个链表交错归并得到1->4->2->5->3;
代码如下：
1 /** 2 * Definition for singly-linked list. 3 * class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * } 10 * } 11 */ 12 public class Solution { 13 private ListNode reverse(ListNode head){ 14 ListNode newHead = null; 15 while(head != null){ 16 ListNode temp = head.next; 17 head.next = newHead; 18 newHead = head; 19 head = temp; 20 } 21 return newHead; 22 } 23 24 private void newMerge(ListNode head1,ListNode head2){ 25 ListNode newHead = new ListNode(0); 26 27 while(head1 != null && head2 != null){ 28 newHead.next = head1; 29 head1 = head1.next; 30 newHead = newHead.next; 31 newHead.next = head2; 32 head2 = head2.next; 33 newHead = newHead.next; 34 } 35 if(head1 != null) 36 newHead.next = head1; 37 if(head2 != null) 38 newHead.next = head2; 39 } 40 private ListNode findMiddle(ListNode head){ 41 ListNode slow = head; 42 ListNode fast = head.next; 43 while(fast != null && fast.next != null) 44 { 45 slow = slow.next; 46 fast = fast.next.next; 47 } 48 return slow; 49 } 50 public void reorderList(ListNode head) { 51 if(head == null || head.next == null) 52 return; 53 54 ListNode mid = findMiddle(head); 55 ListNode tail = reverse(mid.next); 56 mid.next = null; 57 58 newMerge(head, tail); 59 } 60 }
【leetcode】Reorder List,布布扣,bubuko.com
标签：des style blog color io for
原文地址：http://www.cnblogs.com/sunshineatnoon/p/3848999.html
