nlog(n)归并链表

例如:

list             1 3 5 7 9 2 4 6 8 

 

level[20]    20层

 

1.   level[0]   1

2.   level[0]   NULL

      level[1]   1 3 

3    level[0] 5

      level[1] 1 3

4    level[0] NULL

      level[1] NULL

      level[2] 1 3 5 7

5    level[0] 9

      level[1] NULL

      level[2] 1 3 5 7

6    level[0] NULL

      level[1] 2 9

      level[2] 1 3 5 7

7    level[0] 4

      level[1] 2 9

      level[2] 1 3 5 7

8    level[0] NULL

      level[1] NULL

      level[2] NULL

     level[3]  1 2 3 4 5 6 7 9

9   level[0] 8

      level[1] NULL

      level[2] NULL

     level[3]  1 2 3 4 5 6 7 9

 

最后再一次总的归并

1 2 3 4 5 6 7 8 9 

 

class Solution {
public:
	ListNode* level[20];
	ListNode* sortList(ListNode* head) {
		if (head == NULL)return NULL;

		ListNode* cur = head;
		while (cur != NULL)
		{
			int i = 0;
			ListNode* cc = cur;
			cur = cur->next;
			cc->next = NULL;

			while (level[i] != NULL) {
				//归并level[i]和cc
				cc = mar(level[i], cc);
				level[i] = NULL;
				i++;
			}
			level[i] = cc;
		}
		vector<int> lev;
		for (int i = 0; i < 20; i++)
		{
			if (level[i])
				lev.push_back(i);
		}

		for (int i = 1; i < lev.size(); i++)
		{
			level[lev[0]] = mar(level[lev[0]], level[lev[i]]);
		}
		return level[lev[0]];
	}
        //归并a,b
	ListNode* mar(ListNode* a, ListNode* b) {
		ListNode ret(0);
		ListNode* cur = &ret;
		while (a || b)
		{
			if (a && b)
			{
				bool alessb = (a->val < b->val);
				cur->next = alessb ? a : b;
				if (alessb)
					a = a->next;
				else
					b = b->next;

				cur = cur->next;
			}
			else if (a)
			{
				cur->next = a;
				cur = cur->next;

				a = a->next;
			}
			else {
				cur->next = b;
				cur = cur->next;
				b = b->next;
			}
		}
		cur->next = NULL;
		return ret.next;
	}
};