题目：
Sort a linked list using insertion sort.
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) {
ListNode *p1 = head;
ListNode dummy(INT_MIN);
while (p1)
{
ListNode *tmp1 = p1->next;
ListNode *p2 = &dummy;
while ( p2->next )
{
if ( p2->next->val > p1->val )
{
ListNode *tmp2 = p2->next;
p2->next = p1;
p1->next = tmp2;
break;
}
else
{
p2 = p2->next;
}
}
if (!p2->next)
{
p2->next = p1;
p1->next = NULL;
}
p1 = tmp1;
}
return dummy.next;
}
};
tips：
插入排序算法在链表上的实现。
1. 设立一个虚表头dummy，虚表头后面接的就是已经排序好的部分ListNodes
2. 维护一个指针p1，始终指向待插入的ListNode
3. 里层的while循环需要选择插入的具体位置
=============================================
第二次过这道题，一次AC。
/**
* 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) {
ListNode dummpy(INT_MIN);
while ( head )
{
ListNode* tmp = head->next;
ListNode* pre = &dummpy;
ListNode* curr = dummpy.next;
while ( curr )
{
if ( head->val<curr->val)
{
pre->next = head;
head->next = curr;
break;
}
else
{
pre = curr;
curr = curr->next;
}
}
if ( !curr )
{
pre->next = head;
head->next = NULL;
}
head = tmp;
}
return dummpy.next;
}
};