【Insertion Sorted List】cpp

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;
    }
};