Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 思路:合并k路有序链表,可以用最小堆,然后每次送入堆中,取出最小的,并将该链表下一值取出放入 1、使用优先级队列模拟小根堆 /** * Definition for singly-linked list. * struct ListNode { * int va...
Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity. https://oj.leetcode.com/problems/merge-k-sorted-lists/ 思路1(naive):1跟2合并,然后跟3合并,然后跟4合并,一直到k个合并完。 复杂度:1,2合并访问2n个节点,12与3合并访问3n个节点,...,123~k-1与...
Sort a linked list in O(n log n) time using constant space complexity. 本题就是考察的是链表的归并排序。 代码如下: /*class ListNode { int val; ListNode next; ListNode(int x) { val = x; } }*/ public class Solution { public ListNode sortList(ListNode head) { return mergeSort(head);...
Mergeksorted 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 1. 2. 3. 4. 5. 6. 7. 题解: 直接暴力了 classSolution{ public: ListNode*mergeKLists(vector...
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */classSolution{public:ListNode*mergeKLists(vector<ListNode*>&lists){if(lists.empty())returnNULL;vector<int>num;ListNode*temp;for(int i=0...
importheapq# Definition for singly-linked list.# class ListNode:# def __init__(self, x):# self.val = x# self.next = NoneclassSolution:defmergeTwoLists(self,l1:ListNode,l2:ListNode)->ListNode:# 法二:堆。如果使用堆,最原始的办法就是把两个链表的所有节点的值全部放到小顶堆里去,然后重新构...
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题意:归并k个有序的链表。 我的是递归的两路归并 java代码: classMain23{publicListNodemergeKLists(ListNode[]lists){if(lists==null||lists.length==0)returnnull;returnsort(lists,0,lists.length-1)...
977. Squares of a Sorted Array (很像merge sort里的merge）283. Move Zeroes 27. Remove Element ...
举个例子来说比如合并6个链表,那么按照分治法,我们首先分别合并1和4,2和5,3和6。这样下一次只需合并3个链表,我们再合并1和3,最后和2合并就可以了。 总结 关于堆,理解的不是很透彻,可以参考以下两个文章继续学习:http://bubkoo.com/2014/01/14/sort-algorithm/heap-sort/...
# Definitionforsingly-linked list.#classListNode:# def__init__(self,x):# self.val=x # self.next=NoneclassSolution:defmerge(self,node_a,node_b):dummy=ListNode(None)cursor_a,cursor_b,cursor_res=node_a,node_b,dummywhilecursor_a and cursor_b:# 对两个节点的 val 进行判断,直到一方的 nex...