Merge k Sorted Lists

本文介绍了一种使用最小堆实现的高效算法来合并多个已排序的链表,并详细展示了如何利用C++标准模板库中的优先队列实现该算法。

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


/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
bool operator <(const pair<ListNode*,int> &p1, const pair<ListNode*,int> &p2){
    return p1.first->val < p2.first->val;
}

/*假设投针数组为副本,且由其它地方销毁*/
#include<queue>
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode* helpNode = new ListNode(0);
        ListNode* cur = helpNode;

        typedef pair<ListNode*,int> heaptype;
        priority_queue<heaptype,vector<heaptype>,greater<heaptype>> q;//最小队列
        for(int i = 0; i< lists.size(); ++i){
            if(lists[i]){
                q.push(make_pair(lists[i],i));
                lists[i] = lists[i]->next;
            }
        }

        while(!q.empty()){
            ListNode* node = q.top().first;
            int index = q.top().second;
            q.pop();

            if(lists[index]){
                q.push(make_pair(lists[index],index));
                lists[index] = lists[index]->next;
            }

            cur->next = new ListNode(node->val);
            cur= cur->next;
        }
        
        auto ret = helpNode->next;
        delete helpNode;
        return ret;


    }
};

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'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's value is less than the second node'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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