Leetcode -- Insertion Sort List

题目：

Sort a linked list using insertion sort.

分析：

插入排序。

代码1：

/**
* 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 || !head->next) return head;
ListNode* l1 = head;
ListNode* l1_next;
ListNode* last = l1;
ListNode* l2 = head->next;
ListNode* l2_next;

last -> next = NULL;
l1->next = NULL;
while(l2)
{
l1 = head;
l2_next = l2->next;
while(l1)
{
if(last->val <= l2->val)
{
last->next = l2;
last = last->next;
last->next = NULL;
break;
}
if(l1->val > l2->val)
{
l2->next = l1;
head = l2;
break;
}
else
{
if(l1->val <= l2->val && l1->next && l1->next->val > l2->val)
{
l1_next = l1->next;
l1->next = l2;
l2->next = l1_next;
break;
}
else
{
l1 = l1->next;
}
}
}
l2 = l2_next;
}
return head;
}
};

【参考代码：https://leetcode.com/discuss/24663/an-easy-and-clear-way-to-sort-o-1-space】

public ListNode insertionSortList(ListNode head) {
if( head == null ){
return head;
}

ListNode helper = new ListNode(0); //new starter of the sorted list
ListNode cur = head; //the node will be inserted
ListNode pre = helper; //insert node between pre and pre.next
ListNode next = null; //the next node will be inserted
//not the end of input list
while( cur != null ){
next = cur.next;
//find the right place to insert
while( pre.next != null && pre.next.val < cur.val ){
pre = pre.next;
}
//insert between pre and pre.next
cur.next = pre.next;
pre.next = cur;
pre = helper;
cur = next;
}

return helper.next;
}

分析：

思路都是插入排序的思想。代码1比参考代码麻烦的原因是，每次好到插入的地方，就进行插入了。其实，只需要确定插入的地方，然后再进行插入就可以了。没有必要对插入地方进行分类。

其次就是比较的时候，等号的处理。代码1中，是需要考虑这个问题的。