147. Insertion Sort List

题目：

Sort a linked list using insertion sort.

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Linked List Sort

链接： https://leetcode.com/problems/insertion-sort-list/

6/12/2017

46ms, 17%

从head开始遍历，碰到相同或者大的就插入，更新next指针

注意

1. 需要用sortedLength来区分已经排好序的部分，或许不用？

2. 设一个dummy node这样prev部分比较好解决

3. 排好序的部分最后一个元素next = null，这样才能在插入之后不形成环

4. 第27行判断是否刚加入的node是排序部分最大的，是的话将其next = null

5. 外循环时每次先保存next，不然内循环会被覆盖。第14行

 1 public class Solution {
 2     public ListNode insertionSortList(ListNode head) {
 3         if (head == null || head.next == null) {
 4             return head;
 5         }
 6         int sortedLength = 1;
 7         ListNode cur = head.next;
 8         ListNode next;
 9         ListNode dummy = new ListNode(-1);
10         dummy.next = head;
11         head.next = null;
12 
13         while (cur != null) {
14             next = cur.next;
15             ListNode prev = dummy, sorted = dummy.next;
16             int i = 0;
17             for (; i < sortedLength; i++) {
18                 if (sorted.val >= cur.val) {
19                     prev.next = cur;
20                     cur.next = sorted;
21                     break;
22                 } else {
23                     prev = sorted;
24                     sorted = sorted.next;
25                 }
26             }
27             if (i == sortedLength) {
28                 prev.next = cur;
29                 cur.next = null;
30             }
31             sortedLength++;
32             cur = next;
33         }
34         return dummy.next;
35     }
36 }

别人的算法精炼很多，看来sortedLength不需要。做法类似于reverse linkedlist, dummy.next最开始是null，所以之前第3点可以保证。而且可以把所有插入的case统一到一种形式，比我的break和i == sortedLength要好很多

 1 public ListNode insertionSortList(ListNode head) {
 2         if( head == null ){
 3             return head;
 4         }
 5         
 6         ListNode helper = new ListNode(0); //new starter of the sorted list
 7         ListNode cur = head; //the node will be inserted
 8         ListNode pre = helper; //insert node between pre and pre.next
 9         ListNode next = null; //the next node will be inserted
10         //not the end of input list
11         while( cur != null ){
12             next = cur.next;
13             //find the right place to insert
14             while( pre.next != null && pre.next.val < cur.val ){
15                 pre = pre.next;
16             }
17             //insert between pre and pre.next
18             cur.next = pre.next;
19             pre.next = cur;
20             pre = helper;
21             cur = next;
22         }
23         
24         return helper.next;
25     }