2. Design Google search 3. Architect a world-wide video distribution system 4. Build Facebook chat 5. Design News Feed 第一轮system design是一位很客气很nice的国人大哥:有很多台机器,设计一个类似 web crawler的一个系统。然后让我把每台机器上面跑的
62. Unique Paths A robot is located at the top-left corner of amxngrid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below...
design-a-file-sharing-system design-a-food-rating-system design-a-number-container-system design-a-stack-with-increment-operation design-a-text-editor design-add-and-search-words-data-structure design-an-atm-machine design-an-expression-tree-with-evaluate-function design-an-ordered-stre...
The possible movements of chess knight are shown in this diagaram: A chess knight can move as indicated in the chess diagram below: We have a chess knight and a phone pad as shown below, the knight can only stand on a numeric cell (i.e. blue cell). Given an integer n, return how...
【题目】A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram belo...
like Gecko) Chrome/106.0.0.0 Safari/537.36 Edg/106.0.1370.42" version="16.2.6" etag="mVVsJ8HYIxvuaLfj5i_z" type="github"><diagram id="50G2GW0lztJ74-xLJCIO">UzV2zq1wL0osyPDNT0nNUTV2VTV2LsrPL4GwciucU3NyVI0MMlNUjV1UjYwMgFjVyA2HrCFY1qAgsSg1rwSLBiADYTaQg2Y1AA==<...
62.Unique Paths Medium A robot is located at the top-left corner of a m x n grid (marked Start in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked Finish in the ...
LeetCode最近很火,我以前不太知道有这么一个很方便练习算法的网站,直到大概数周前同事和我说起,正好我老婆要找工作,而根据同事的理论,LeetCode的题目是必须攻破的第一道关卡。我虽说又不找工作,但是纯粹拿来练手和学习,觉得很多题目都挺有趣的。现在已经做了三分之一,我会把我的解答分几次放上来。这里是第一...
System.out.println(isPalindrome(s)); } // 原来递归对于太长字符串会StackOverflow!改成迭代通过 public static boolean isPalindrome(String s) { if(s==null || s.length()==0 || s.length()==1){ return true; } int i = 0, j = s.length()-1; ...
0631 Design Excel Sum Formula 31.5% Hard 0632 Smallest Range Covering Elements from K Lists Go 52.4% Hard 0633 Sum of Square Numbers Go 32.2% Easy 0634 Find the Derangement of An Array 40.1% Medium 0635 Design Log Storage System 58.6% Medium 0636 Exclusive Time of Functions Go 52.0...