3. 合并K个排序链表

Posted 2020-11-13 zouma

一、进行k-1次的2路合并，虽然排序次数过多，出现重复排序，但是不得不说 使用迭代的合并内存消耗比递归小很多。O（nk）的时间复杂度，O（1）的空间复杂度

 1 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
 2     if (l1 == nullptr)
 3         return l2;
 4     else if (l2 == nullptr)
 5         return l1;
 6     ListNode* temp1 = l1;
 7     ListNode* temp2 = l2;
 8     ListNode* next1 = nullptr;
 9     ListNode* next2 = nullptr;
10     while (temp1 != nullptr&&temp2 != nullptr)
11     {
12         if (temp1->val <= temp2->val && temp1->next)
13         {
14             if (temp1->next->val > temp2->val)
15             {
16                 next2 = temp2->next;
17                 temp2->next = temp1->next;
18                 temp1->next = temp2;
19                 temp2 = next2;
20                 temp1 = temp1->next;
21             }
22             else if (temp1->next->val <= temp2->val)
23                 temp1 = temp1->next;
24         }
25         else if (temp1->val <= temp2->val && !temp1->next)
26         {
27             temp1->next = temp2;
28             break;
29         }
30         else if (temp1->val > temp2->val)
31         {
32             next2 = temp2->next;
33             temp2->next = temp1;
34             temp1 = temp2;
35             temp2 = next2;
36             l1 = temp1;    //bug
37         }
38     }
39     return l1;
40 }
41 ListNode* mergeKLists(vector<ListNode*>& lists)
42 {
43     int size = lists.size();
44     if (size == 0)
45         return nullptr;
46     else if (size == 1)
47         return lists[0];
48     ListNode* res = mergeTwoLists(lists[0], lists[1]);
49     for (int i = 2; i < size; ++i)
50     {
51         res = mergeTwoLists(res, lists[i]);
52     }
53     return res;
54 }

二、分治（题解里是这么说的？），就是两两配对，O（nlogk）的时间复杂度，O（1）的空间复杂度

 1 ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
 2     if (l1 == nullptr)
 3         return l2;
 4     else if (l2 == nullptr)
 5         return l1;
 6     ListNode* temp1 = l1;
 7     ListNode* temp2 = l2;
 8     ListNode* next1 = nullptr;
 9     ListNode* next2 = nullptr;
10     while (temp1 != nullptr&&temp2 != nullptr)
11     {
12         if (temp1->val <= temp2->val && temp1->next)
13         {
14             if (temp1->next->val > temp2->val)
15             {
16                 next2 = temp2->next;
17                 temp2->next = temp1->next;
18                 temp1->next = temp2;
19                 temp2 = next2;
20                 temp1 = temp1->next;
21             }
22             else if (temp1->next->val <= temp2->val)
23                 temp1 = temp1->next;
24         }
25         else if (temp1->val <= temp2->val && !temp1->next)
26         {
27             temp1->next = temp2;
28             break;
29         }
30         else if (temp1->val > temp2->val)
31         {
32             next2 = temp2->next;
33             temp2->next = temp1;
34             temp1 = temp2;
35             temp2 = next2;
36             l1 = temp1;    //bug
37         }
38     }
39     return l1;
40 }
41 ListNode* mergeKLists(vector<ListNode*>& lists)
42 {
43     int size = lists.size();
44     if (size == 0)
45         return nullptr;
46     else if (size == 1)
47         return lists[0];
48     int k = size;
49 
50     int    distance = 1;
51     while (distance<size)
52     {
53         for (int i = 0; i < size; i=i+distance*2)
54         {
55             if(i+distance<size)    //bug
56                 lists[i] = mergeTwoLists(lists[i], lists[i + distance]);
57         }
58         distance *= 2;
59     }
60     return lists[0];
61 }