首页 > 技术文章 > 23. Merge k Sorted Lists

Zzz-y 2018-05-20 15:43 原文

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

合并k个有序链表。

用归并排序就好了。

 1 class Solution {
 2 public:
 3     ListNode* merge(vector<ListNode*>& lists, int low, int high) {
 4         ListNode* left = nullptr;
 5         ListNode* right = nullptr;
 6         if (low < high - 1) {
 7             int mid = low + (high - low) / 2;
 8             left = merge(lists, low, mid);
 9             right = merge(lists, mid+1, high);
10         }
11         else {
12             left = lists[low];
13             right = low == high? nullptr: lists[high];
14         }
15         int a, b; 
16         ListNode *res = new ListNode(0);
17         ListNode *curr = res;
18         while (left || right) {
19             a = left? left->val: INT_MAX;
20             b = right? right->val: INT_MAX;
21             if (a < b) {
22                 curr->next = left;
23                 left = left->next;
24                 curr = curr->next;
25             }
26             else {
27                 curr->next = right;
28                 right = right->next;
29                 curr = curr->next;
30             }
31         }
32         curr = res->next;
33         delete res;
34         return curr;
35     }
36     ListNode* mergeKLists(vector<ListNode*>& lists) {
37         ListNode *head = nullptr;
38         int low = 0;
39         int high = lists.size() - 1;
40         if (low <= high)
41             head = merge(lists, low, high);
42         return head;
43     }
44 };

 

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