LC.143.Reorder List

https://leetcode.com/problems/reorder-list/description/
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}.
time: o(n) space: o(1)


 1 public void reorderList(ListNode head) {
 2         if (head == null || head.next == null) return;
 3         ListNode mid = null;
 4         ListNode l1 = head;
 5         // step 1: find the mid
 6         mid = findMid(head);
 7         ListNode l2 = mid.next ;
 8         //重要!断开中点和后一段
 9         mid.next = null;
10         //step 2: reverse the 2nd half
11         ListNode l2Reverse = reverse(l2);
12         //step 3: merge with the 1st with the 2nd
13         merge(l1, l2Reverse);
14     }
15 
16     //去中点
17     private ListNode findMid(ListNode head){
18         ListNode fast = head, slow = head;
19         while (fast != null && fast.next != null && fast.next.next != null) {
20             fast = fast.next.next;
21             slow = slow.next;
22         }
23         return slow ;
24     }
25 
26     //reverse
27     private ListNode reverse(ListNode head) {
28         if (head == null || head.next == null) return head;
29         ListNode pre = null;
30         ListNode curr = head;
31         while (curr != null) {
32             ListNode temp = curr.next;
33             curr.next = pre;
34             pre = curr;
35             curr = temp;
36         }
37         return pre;
38     }
39 
40     //合并
41     private ListNode merge(ListNode head1, ListNode head2) {
42         ListNode curr1 = head1, curr2 = head2, temp1 = null, temp2 = null;
43         while (curr1.next != null && curr2 != null) {
44             temp1 = curr1.next;
45             temp2 = curr2.next;
46             curr1.next = curr2;
47             curr2.next = temp1;
48             curr1 = temp1;
49             curr2 = temp2;
50         }
51         if (curr2 != null) {
52             curr1.next = curr2;
53             curr2.next = null;
54         }
55         else {
56             curr1.next = null;
57         }
58         return head1;
59     }

 

posted @ 2018-02-25 23:51  davidnyc  阅读(103)  评论(0编辑  收藏  举报