ListNode* head2 = walker->next; walker->next =NULL;//非常重要,将链表断开,一分为二head1 =sortList(head1); head2 =sortList(head2);returnmergeTwoLists(head1, head2); }ListNode *mergeTwoLists(ListNode *l1, ListNode *l2){if(l1 ==NULL)returnl2;elseif(l2 ==NULL)returnl1; ListNode *R...
三刷,把取中间节点的方法提取了出来。 /*** Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * }*/publicclassSolution {publicListNode sortList(ListNode head) {if(head ==null|| head.next ==null) {returnhead; ...
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);...
*fast=head,*pre=head;while(fast&&fast->next){pre=slow;slow=slow->next;fast=fast->next->next;}pre->next=NULL;returnmergeTwoLists(sortList(head),sortList(slow));}ListNode*mergeTwoLists(ListNode*l1,ListNode*l2){if(!l1)returnl2;if(!l2)returnl1;ListNode*dummy=new...
java list集合sort java list集合左连接,集合分为单列集合和双列集合,单列集合的顶级接口是Collection,双列集合的顶级接口是Mapcollection--1.list接口:存储数据的结构:堆栈:先进后出,队列:先进先出,数组:查询快,增删慢,链表:查询慢,增删快。特点:有序,拿出
首先用快慢指针的方法找到链表中间节点,然后递归的对两个子链表排序,把两个排好序的子链表合并成一条有序的链表。归并排序应该算是链表排序最佳的选择了,保证了最好和最坏时间复杂度都是nlogn,而且它在数组排序中广受诟病的空间复杂度在链表排序中也从O(n)降到了O(1) ...
linked list, using constant space complexity. """ def sortList(self, head): # write your code here def merge(list1,list2): if list1 == None: return list2 if list2 == None: return list1 head = None if list1.val < list2.val: head = list1 list1 = list1.next else: head =...
def sortList(self, head): """ :type head: ListNode :rtype: ListNode """ # 如果为空或者只有一个元素则直接返回 if not head or not head.next: return head # count表示节点个数 count, node = 0, head # 获取节点的个数 while node: ...
Sort a linked list inO(nlogn) time using constant space complexity. 解题思路:看到这个时间复杂度就能想到归并排序或者快速排序算法,这里我就用相对简单的归并排序解决,解决思路和数组的归并排序一致。 func sortList(head *ListNode) *ListNode { if head == nil || head.Next == nil { ...
LinkedHashMap yes yes* yes key HashBidiMap no no no key* TreeBidiMap yes yes* yes key* Trees RedBlackTree yes yes* no key AVLTree yes yes* no key BTree yes yes* no key BinaryHeap yes yes* no index *reversible *bidirectional Lists A list is a data structure that stores values and...