[LintCode] 599 Insert into a Cyclic Sorted List 解题报告

Description
Given a node from a cyclic linked list which has been sorted, write a function to insert a value into the list such that it remains a cyclic sorted list. The given node can be any single node in the list. Return the inserted new node.

Notice
3->5->1 is a cyclic list, so 3 is next node of 1.
3->5->1 is same with 5->1->3

Example
Given a list, and insert a value 4:
3->5->1
Return 5->1->3->4

5/18/2017

算法班，未经验证

先找到list的头，再来循环找插入点。

 1 public class Solution {
 2     /**
 3      * @param node a list node in the list
 4      * @param x an integer
 5      * @return the inserted new list node
 6      */
 7     public ListNode insert(ListNode node, int x) {
 8         // Write your code here
 9
10         ListNode newNode = new ListNode(x);
11         if (node == null) {
12             newNode.next = newNode;
13             return newNode;
14         } else if (node.next == node) {
15             node.next = newNode;
16             newNode.next = node;
17             return node;
18         }
19         ListNode maxNode = node;
20
21         while (maxNode.next.val > maxNode.val) {
22             maxNode = maxNode.next;
23         }
24         if (maxNode.val < newNode.val || maxNode.next.val > newNode.val) {
25             newNode.next = maxNode.next;
26             maxNode.next = newNode;
27             return newNode;
28         }
29
30         ListNode node = maxNode.next;
31         while (node.next.val < newNode.val) {
32             node = node.next;
33         }
34         newNode.next = node.next;
35         node.next = newNode;
36         return newNode;
37     }
38 }

但是实际上可以在单纯遍历的时候找插入点

别人的答案

 1 public class Solution {
 2     /**
 3      * @param node a list node in the list
 4      * @param x an integer
 5      * @return the inserted new list node
 6      */
 7     public ListNode insert(ListNode node, int x) {
 8         // Write your code here
 9
10         if (node == null) {
11             node = new ListNode(x);
12             node.next = node;
13             return node;
14         }
15
16         ListNode head = node;
17         while (node != null && node.next != null) {
18             if (node.val < node.next.val) {
19                 if (node.val <= x && x <= node.next.val) {
20                     insertNode(node, x);
21                     break;
22                 }
23             }
24             else if (node.val > node.next.val) {
25                 if (x > node.val || x < node.next.val) {
26                     insertNode(node, x);
27                     break;
28                 }
29             }
30             else { // node.val == node.next.val
31                 if (node.next == head) {
32                     insertNode(node, x);
33                     break;
34                 }
35             }
36             node = node.next;
37         }
38
39         return head;
40     }
41
42     public void insertNode(ListNode node, int x) {
43         ListNode newNode = new ListNode(x);
44         newNode.next = node.next;
45         node.next = newNode;
46     }
47 }