[LeetCode]Sort List 题目描述:Sort a linked list in O(n log n) time using constant space complexity.解题方案:题目要求的时间复杂度是 O(n log n),常数级空间复杂度。所以这里用了归并排序,归并排序在数组上操作比较方便,但是这里要排序的是链表。我们用到两个指针将链表一份为二,一个指针q每次前进两...
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...
leetcode -- Sort List https://leetcode.com/problems/sort-list/ 这里又复习了一遍merge list的操作。用归并排序就行 AI检测代码解析 class Solution(object): def mergeList(self, head1, head2): if not head1: return head2 if not head2: return head1 dummy = ListNode(0) p = dummy while hea...
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* sortList(ListNode* head) { if(head==NULL||head->next==NULL)//空链表||只有一个节点的链表 return...
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 { ...
用Python 时间也算不短了,但总感觉自己在用写 C++ 代码的思维写 Python,没有真正用到其作为脚本语言的优势。之前刷 LeetCode 时,自己的 Python 代码总是很长,很像披着 Python 外衣的 C++ 代码(放在这里( https://github.com/xuelangZF/LeetCode ),不断重构中)。
:pencil2: 算法相关知识储备 LeetCode with Python :books:. Contribute to tomorrow1029/leetCode development by creating an account on GitHub.
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) ...
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...