https://oj.leetcode.com/discuss/3577/i-use-quick-sort-to-sort-the-list-but-why-i-get-time-limited View Code GITHUB: https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/list/SortList.java
For each round, the iteration is O(n). And for merge sort, we totally need run the iteration for lg(n) round(the height of recursion tree). So, the total time complexity is O(nlgn). Maybe someone can share a brief implementation. My current code is a bit fat. 1:ListNode*sortList...
Can you solve this real interview question? Sort List - Given the head of a linked list, return the list after sorting it in ascending order. Example 1: [https://assets.leetcode.com/uploads/2020/09/14/sort_list_1.jpg] Input: head = [4,2,1,3] Outpu
classSolution{public:ListNode*sortList(ListNode*p){if(p==NULL||p->next==NULL)returnp;//增加一个头节点,避免合并时讨论rear为空的情况.ListNode*head=newListNode(-1),*q=head;head->next=p;intcnt=0;while(p){++cnt;p=p->next;if(cnt%2==0)q=q->next;}p=q->next,q->next=NULL;//递归...
【leetcode】sort list(python) 链表的归并排序 超时的代码 class Solution: def merge(self, head1, head2): if head1 == None: return head2 if head2 == None: return head1 # head1 and head2 point to the same link list if head1 == head2:...
classSolution:defmissingNumber(self,nums:List[int])->int:nums.sort()ifnums[-1]!=len(nums):returnlen(nums)elifnums[0]!=0:return0foriinrange(1,len(nums)):#从1开始检查,把0作为nums[i-1]+1的起点expected_num=nums[i-1]+1ifnums[i]!=expected_num:returnexpected_num ...
Loading...leetcode.com/problems/course-schedule-iii/discuss/104847/Python-Straightforward-with-Explanation 下面是我的解答,太慢,可以用来理解解题过程,但不用参考。 importheapqclassSolution:defscheduleCourse(self,courses:List[List[int]])->int:# 优先选早结课的课程,无论这门课花的时间长短,都会是最优...
class Solution { public ListNode sortList(ListNode head) { if(head==null||head.next==null) return head; ListNode dummy = new ListNode(0); ListNode pre = dummy; dummy.next = head; ListNode slow = head; ListNode fast = head; //使用快慢指针切分链表 ...
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 { ...
代码如下：class Solution(object):def reorderLogFiles(self, logs):""":type logs: List[str]:rtype: List[str]"""l1=[]l2=[]for l in logs:if l[-1].isalpha():l1.append(l)else:l2.append(l)l1.sort(key=lambda x:(x[x.index(' ')+1:],x[:x.index(' ')]))