java代码: 1/**2* Definition for singly-linked list.3* class ListNode {4* int val;5* ListNode next;6* ListNode(int x) {7* val = x;8* next = null;9* }10* }11*/12publicclassSolution {13publicListNode sortList(ListN
1/**2* Definition for singly-linked list.3* struct ListNode {4* int val;5* ListNode *next;6* ListNode(int x) : val(x), next(NULL) {}7* };8*/9classSolution {10//需要注意,链表版的merge()要充分发挥链表的特性,不需要想顺序表那样进行复制11//归并的基本思想是不断将两条待排序链上的...
LeetCode:Sort List Problem: Sort a linked list inO(nlogn) time using constant space complexity. 解题思路: 首先,时间复杂度能达到O(nlgn)的排序算法,常见的有3种:堆排序、归并排序和高速排序, 而对于链表,用堆排序显然不太可能,所以,我们可用归并或者是快排.因为合并两个链表,仅仅用 改动对应的指针,所以...
leetcode 148. Sort List 链表归并排序 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(Lis...
*/classSolution{publicListNodesortList(ListNode head){if(head==null||head.next==null)returnhead;// 1. find mid of listListNode fast=head,slow=head,partHead=slow;while(fast!=null&&fast.next!=null){partHead=slow;fast=fast.next.next;slow=slow.next;}// 2. cut list as binarypartHead.nex...
Code Issues Pull requests GoDS (Go Data Structures) - Sets, Lists, Stacks, Maps, Trees, Queues, and much more gomapgolangsetlisttreedata-structureavl-treestackqueueiteratorsortred-black-treeenumerablebinary-heapb-tree UpdatedMar 12, 2025 ...
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package leetcode /** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func insertionSortList(head *ListNode) *ListNode { if head == nil { return head } newHead := &ListNode{Val: 0, Next: nil} // 这里初始化不要直接指向 head,为了...
1.Heap表示方法 满足以下性质的二叉树Binary Tree可以成为Binary Heap: Complete Tree:所有的层都是完全的,除了最后一层,且最后一层的叶子靠左。 Min Heap or Max Heap:根节点是最大值或者最小值,而且这个性质对于任何递归得到的子树都成立。 Binary Heap通常使用array表示: 根节点在array[0]; array[(i-1)//...
代码# Go packageleetcode/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */funcsortList(head*ListNode)*ListNode{length:=0cur:=headforcur!=nil{length++cur=cur.Next}iflength<=1{returnhead}middleNode:=middleNode(head)cur=middleNode.Next ...