标签：des style blog http color os io ar 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}.
这是一道比较综合的链表操作的题目，要按照题目要求给链表重新连接成要求的结果。其实理清思路也比较简单，分三步完成：（1）将链表切成两半，也就是找到中点，然后截成两条链表；（2）将后面一条链表进行reverse操作，就是反转过来；（3）将两条链表按顺序依次merge起来。
这几个操作都是我们曾经接触过的操作了，第一步找中点就是用runner technique方法，一个两倍速跑，一个一倍速跑，知道快的碰到链表尾部，慢的就正好停在中点了。第二步是比较常见的reverse操作，在Reverse Nodes in k-Group也有用到了，一般就是一个个的翻转过来即可。第三步是一个merge操作，做法类似于Sort List中的merge
接下来看看时间复杂度，第一步扫描链表一遍，是O(n)，第二步对半条链表做一次反转，也是O(n)，第三部对两条半链表进行合并，也是一遍O(n)。所以总的时间复杂度还是O(n)，由于过程中没有用到额外空间，所以空间复杂度O(1)。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 public void reorderList(ListNode head) { 14 if (head == null || head.next == null) return; 15 ListNode dummy = new ListNode(-1); 16 dummy.next = head; 17 ListNode current = dummy; 18 ListNode runner = dummy; 19 while (runner.next != null && runner.next.next != null) { 20 current = current.next; 21 runner = runner.next.next; 22 } 23 ListNode half = current.next; 24 current.next = null; 25 26 /*reverse the second Linked List*/ 27 if (runner.next != null) runner = runner.next; //make sure the runner pointer points to the end of the the second Linked List 28 ListNode dummy2 = new ListNode(-2); 29 dummy2.next = half;//create another dummy node whose next points to the head of the second Linked List 30 while (dummy2.next != runner) { 31 ListNode dnext = dummy2.next.next; 32 ListNode rnext = runner.next; 33 dummy2.next.next = rnext; 34 runner.next = dummy2.next; 35 dummy2.next = dnext; 36 } 37 38 /*merge the two Linked List together*/ 39 ListNode header1 = dummy; 40 ListNode header2 = dummy2; 41 while (header1.next != null && header2.next != null) { 42 ListNode store = header1.next.next; 43 ListNode merge = new ListNode(header2.next.val); 44 header1.next.next = merge; 45 merge.next = store; 46 header1 = header1.next.next; 47 header2 = header2.next; 48 } 49 if (header2.next != null) { 50 header1.next = header2.next; 51 } 52 } 53 }
标签：des style blog http color os io ar for
原文地址：http://www.cnblogs.com/EdwardLiu/p/3960290.html
