23 Merge k Sorted Lists

本文介绍了一种解决LeetCode中难度为hard的题目——合并k个有序链表的方法。作者通过C++实现,详细阐述了思路及过程中遇到的错误,如未考虑容器为空的情况。通过两两归并策略,最终达到O(m * n)的时间复杂度和O(1)的空间复杂度。

归并多组有序的链表

刚刚在LeetCode上刷了一道题目,显示是hard级别的,但是自己居然编译痛过之后一遍就AC了,想想挺开心的,就写篇博客分析一下这道题目,并且分析这道题目的时间复杂度等
先看题目:

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
以下是C++的实现入口:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
    }
}

题目中给定一个容器,里面存储的是每一个链表的第一个节点,所以我们首先需要获得所有的开始节点,然后两两归并。我的思路是将第一个链表作为head,以后开始顺序遍历容器,每一次取出容器内的一个链表,将它和之前已经合并了的链表合并,直至容器为空,然后返回head就好了。
值得一提的是在我第一遍提交的时候出现了一个报错:reference binding to null pointer of type ‘st

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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