Sort a linked list using insertion sort.
我原本的想法是用额外的空间拷贝每一个节点，建立了一个新的sorted的LinkedList, 后来看到别人的做法不用建立新的LinkedList, 直接以原有List上的节点组成新的sorted的LinkedList。我之前想这样做会扰乱遍历的顺序，但是其实sorted的list和unsorted list是完全分开互不影响的。先还是给sorted list建立一个dummy node做前置节点， 每次取unsorted list里面的一个节点，记录下下一跳位置，然后把这个点插入到sorted list对应位置上（pre.next指到null或者pre.next。val更大）。
Insert Sort的做法相当于是每次从原来的List里面删除头节点，再把这个头节点插入到新的List里相应的位置。这个新List全由原来的节点组成，只是变换了顺序。
时间复杂度是插入排序算法的O(n^2)，空间复杂度是O(1)。
第二遍做法：
1 public class Solution {
2 public ListNode insertionSortList(ListNode head) {
3 ListNode dummy = new ListNode(-1);
4 ListNode cursor = dummy;
5 while (head != null) {
6 ListNode next = head.next;
7 while (cursor.next!=null && cursor.next.val<=head.val) {
8 cursor = cursor.next;
9 }
10 head.next = cursor.next;
11 cursor.next = head;
12 head = next;
13 cursor = dummy;
14 }
15 return dummy.next;
16 }
17 }
第一遍做法：
1 /**
2 * Definition for singly-linked list.
3 * public 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 ListNode insertionSortList(ListNode head) {
14 if(head == null)
15 return null;
16 ListNode helper = new ListNode(0);
17 ListNode pre = helper;
18 ListNode cur = head;
19 while(cur!=null)
20 {
21 ListNode next = cur.next;
22 pre = helper;
23 while(pre.next!=null && pre.next.val<=cur.val)
24 {
25 pre = pre.next;
26 }
27 cur.next = pre.next;
28 pre.next = cur;
29 cur = next;
30 }
31 return helper.next;
32 }
33 }
上面程序注释如下：
1 public ListNode insertionSortList(ListNode head) {
2 if(head == null)
3 return null;
4 ListNode helper = new ListNode(0);//helper is the dummy head for the result sorted LinkedList
5 ListNode pre = helper; //pre is the pointer to find the suitable place to plug in the current node
6 ListNode cur = head; //cur is current node, the node to be plugged in the sorted list
7 while(cur!=null)
8 {
9 ListNode next = cur.next; //keep a record of the current node's next
10 pre = helper; //after one search, put pre back to its original place
11 while(pre.next!=null && pre.next.val<=cur.val) //use pre to traverse the sorted list to find the suitable place to plug in
12 {
13 pre = pre.next;
14 }
15 cur.next = pre.next;// plug in current node to the right place
16 pre.next = cur;
17 cur = next; //go on to deal with the next node in the unsorted list
18 }
19 return helper.next;
20 }
来源：https://www.cnblogs.com/EdwardLiu/p/3976821.html