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摘要: Given two sparse matrices A and B, return the result of AB. You may assume that A's column number is equal to B's row number. Example: 阅读全文
posted @ 2018-01-28 03:06 逸朵 阅读(88) 评论(0) 推荐(0) 编辑
摘要: Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference room 阅读全文
posted @ 2018-01-28 02:05 逸朵 阅读(112) 评论(0) 推荐(0) 编辑
摘要: Given two binary strings, return their sum (also a binary string). For example,a = "11"b = "1"Return "100". 阅读全文
posted @ 2018-01-26 09:01 逸朵 阅读(81) 评论(0) 推荐(0) 编辑
摘要: Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results. Note: The input string may cont 阅读全文
posted @ 2018-01-26 08:18 逸朵 阅读(141) 评论(0) 推荐(0) 编辑
摘要: Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could 阅读全文
posted @ 2018-01-26 03:54 逸朵 阅读(95) 评论(0) 推荐(0) 编辑
摘要: I have read many posts on SO and the web regarding the keywords in my question title and learned a lot from them. Some of the questions I read are rel 阅读全文
posted @ 2018-01-22 12:39 逸朵 阅读(290) 评论(0) 推荐(0) 编辑
摘要: Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are n 阅读全文
posted @ 2018-01-22 10:01 逸朵 阅读(106) 评论(0) 推荐(0) 编辑
摘要: Given an unsorted integer array, find the first missing positive integer. For example,Given [1,2,0] return 3,and [3,4,-1,1] return 2. Your algorithm s 阅读全文
posted @ 2018-01-22 09:50 逸朵 阅读(111) 评论(0) 推荐(0) 编辑
摘要: In a nutshell, ZooKeeper helps you build distributed applications. How it works You may describe ZooKeeper as a replicated synchronization service wit 阅读全文
posted @ 2018-01-21 11:48 逸朵 阅读(194) 评论(0) 推荐(0) 编辑
摘要: Kafka in a Nutshell Posted on September 25, 2015 • kafka Kafka in a Nutshell Posted on September 25, 2015 • kafka Kafka in a Nutshell Posted on Septem 阅读全文
posted @ 2018-01-21 11:05 逸朵 阅读(211) 评论(0) 推荐(0) 编辑
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