6.5-9 merge k sorted lists into one sorted list with O(n lg k)

本文提供了一种使用最大堆实现的高效算法,能够在O(nlgk)时间内将k个有序列表合并成一个有序列表。该算法通过维护一个包含k个列表头部元素的最大堆,并在每次取出最大元素后更新堆来完成合并。

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Give an O(n lg k)-time algorithm to merge k sorted lists into one sorted list,

where n is the total number of elements in all the input lists. (Hint: Use a minheap

for k-way merging.)


Solution

1. For each sorted list, maintain a 0 initilized array of k elements each representing the index of the current element to be dealt.

2. Build a k-element max heap, the elements are from the cur head of the k sorted list.

3. Each time, get the max from the heap, (lg k) and insert a new element from which the max of the heap is from before the delete.(lg k ) These two steps can be combined to consume only a single lg k, instead of a double lg k -> Record the max, replace the max with the new element, and then let it sink downward.

If all elements from a certain array has dealt, use minux INT_MAX as 

4. Put the max at the final array of element N.

5. repeat 3 and 4, until the number of dealt elements is N.


From: http://blog.youkuaiyun.com/cheetach119/article/details/9353035



To merge k sorted linked lists, one approach is to repeatedly merge two of the linked lists until all k lists have been merged into one. We can use a priority queue to keep track of the minimum element across all k linked lists at any given time. Here&#39;s the code to implement this idea: ``` struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; // Custom comparator for the priority queue struct CompareNode { bool operator()(const ListNode* node1, const ListNode* node2) const { return node1->val > node2->val; } }; ListNode* mergeKLists(vector<ListNode*>& lists) { priority_queue<ListNode*, vector<ListNode*>, CompareNode> pq; for (ListNode* list : lists) { if (list) { pq.push(list); } } ListNode* dummy = new ListNode(-1); ListNode* curr = dummy; while (!pq.empty()) { ListNode* node = pq.top(); pq.pop(); curr->next = node; curr = curr->next; if (node->next) { pq.push(node->next); } } return dummy->next; } ``` We start by initializing a priority queue with all the head nodes of the k linked lists. We use a custom comparator that compares the values of two nodes and returns true if the first node&#39;s value is less than the second node&#39;s value. We then create a dummy node to serve as the head of the merged linked list, and a current node to keep track of the last node in the merged linked list. We repeatedly pop the minimum node from the priority queue and append it to the merged linked list. If the popped node has a next node, we push it onto the priority queue. Once the priority queue is empty, we return the head of the merged linked list. Note that this implementation has a time complexity of O(n log k), where n is the total number of nodes across all k linked lists, and a space complexity of O(k).
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