题目：Sort a linked list using insertion sort.
即使用插入排序对链表进行排序。
思路分析：
插入排序思想见《排序（一）：直接插入排序 》
C++参考代码：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
public:
ListNode *insertionSortList(ListNode *head)
{
if (!head) return nullptr;
//previous记住当前节点的前一个节点（因为单链表进行插入操作的时候必须知道当前节点的前一个节点）
ListNode *previous = head;
//current节点是要往前面已经排好序的节点中进行插入的节点
ListNode *current = head->next;
while (current)
{
//small和big指针用于记录比current小和比current大的指针，small和big相邻
ListNode *small = nullptr;
ListNode *big = head;
while (big->val < current->val)
{
small = big;
big = big->next;
}
//如果big!=current说明current节点应该插入到small和big节点之间
if (big != current)
{
previous->next = current->next;
if (small) small->next = current;
else head = current;
current->next = big;
}
//如果big=current这时previous指向下一个节点就OK了
else
{
previous = previous->next;
}
//最后current节点指向previous下一个节点即current向前走一步
current = previous->next;
}
return
Java参考代码：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution
public ListNode insertionSortList(ListNode head) {
if (head == null) return null;
ListNode previous = head;
ListNode current = head.next;
while (current != null) {
ListNode small = null;
ListNode big = head;
//找出介于current的big和small元素，big和small相邻
while (big.val < current.val) {
small = big;
big = big.next;
}
//如果big!=current将current插入到small和big之间
if (big != current) {
previous.next = current.next;
if (small != null) small.next = current;
else head = current;
current.next = big;
} else {
previous = previous.next;
}
current = previous.next;
}
return
