Challenge O(n log n) time and O(1) extra space. Note 方法上没太多难点,先按所有区间的起点排序,然后用pre和cur两个指针,如果有交集进行merge操作,否则pre向后移动。由于要求O(1)的space,就对原数组直接进行操作了。 时间复杂度O(nlogn)是Collections.sort()的时间。for循环是O(n)。 这道题有两个点...
2181. Merge Nodes in Between Zeros Medium Topics Companies Hint You are given theheadof a linked list, which contains a series of integersseparatedby0's. Thebeginningandendof the linked list will haveNode.val == 0. Foreverytwo consecutive0's,mergeall the nodes lying in between them into ...
It's the LeetCode Challenge. Created by @caio-andres & @matheusaudibert & @enzowxl to improve our structure algorithm skill using https://leetcode.com/. - Merge pull request #4 from Ancelotti-Dev/main · caio-andres/leetcode-challenge@bf56c33
- [ ] [LeetCode - 146 LRU Cache (C++) (video)](https://www.youtube.com/watch?v=8-FZRAjR7qU) - [ ] CPU cache: - [ ] [MIT 6.004 L15: The Memory Hierarchy (video)](https://www.youtube.com/watch?v=vjYF_fAZI5E&list=PLrRW1w6CGAcXbMtDFj205vALOGmiRc82-&index=24) - [...
Challenge O(nlogn) time and O(1) extra space. 虽然它本来无序,但我们可以将它根据 start 值的大小进行排序,可以使用 Collections 类中的 sort 方法对 List 进行排序。排完之后,就可以对集合内的区间进行合并了。 申请一个新的集合,再用一个循环,将排好序的区间两两比较,如果无需合并,则将前者加入新的集...