[LeetCode] Merge k Sorted Lists

最新推荐文章于 2024-04-03 21:24:25 发布

转载最新推荐文章于 2024-04-03 21:24:25 发布 · 49 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4623367.html

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#开发工具

本文介绍了一种高效算法来合并多个已排序的链表。该算法通过不断合并未合并的链表直到只剩下一个链表的方式解决问题，并避免了顺序合并导致的时间效率低下问题。

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Well, the idea of this problem is actually very sample --- keep merging the unmerged lists in lists until there is exactly one list remained. However, do not merge lists[0] with lists[1], lists[2], ..., lists[n - 1] sequentially since that will cause TLE. You may merge the first two lists and then push it back to lists and erase the first two to avoid repeated merging, or use some variables to control the merging process, as the following code.

 1 class Solution {
 2 public:
 3     ListNode* mergeKLists(vector<ListNode*>& lists) {
 4         if (lists.empty()) return NULL;
 5         int n = lists.size();
 6         while (n > 1) {
 7             for (int i = 0; i < n / 2; i++)
 8                 lists[i] = mergeLists(lists[i], lists[n - 1 - i]);
 9             n = (n + 1) / 2;
10         }
11         return lists[0];
12     }
13 private:
14     ListNode* mergeLists(ListNode* head1, ListNode* head2) {
15         ListNode* new_head = new ListNode(0);
16         ListNode* run = new_head;
17         ListNode* run1 = head1;
18         ListNode* run2 = head2;
19         while (run1 && run2) {
20             if (run1 -> val <= run2 -> val) {
21                 run -> next = run1;
22                 run1 = run1 -> next;
23             }
24             else {
25                 run -> next = run2;
26                 run2 = run2 -> next;
27             }
28             run = run -> next;
29         }
30         run -> next = run1 ? run1 : run2;
31         return new_head -> next;
32     }
33 };