字节跳动挂经

上周面的，应该是挂了所以没联系，分享一下挂经。想吐槽一下他家hr，首先是很难联系上，有任何问题发邮件过去问没回音。其次面试给我安排了两轮也不在邮件里说明，而且两轮之间还间隔了十分钟左右，还好我当时没关页面面试官上线我才知道还有一轮。

第一轮：给一个字符串，要求你把里面重复的字符删除，保证每个字符只能出现一次，如果结果有多种可能性的话，要

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充内容 (2026-06-30 12:48 +08:00):
时隔六年，又要面tiktok，回来看发现这两题都是leetcode的原题，当年真的是自己学术不精。。。

第一题是刷题网316

qiaofengmarco 发表于 2020-5-14 06:26
老哥你考虑一下 cbacbc 这个情况

他的解法好像没什么问题吧, 用他的方法解cbacbc的话先是 cba -> bac -> acb -> abc

ozox 发表于 2020/05/14 07:13:52
if input is: cbacbc

1. first, result = cba
2. next, 'c' ...

Not work for cdacd

分人吧
我之前面的时候hr回复很快 描述也很清楚
可惜我碰到了不咋地的面试官 虽然水平应该很好 但不喜欢讨论 态度很冷 也不care你的回答

补充内容 (2020-5-12 16:11):
第一题想到一种O(n)解法
第二题没什么思路 不知道能不能dfs

同面了两轮面试官人很好交流很顺畅，题目很快做出来followup也接住了，但是还是挂了。。。不知道为什么，收到的邮件是我的work experience太少？？？

First problem:

1. Use a map, key is the char, value is a node of a doubly linked list (or just the location of the char in the input array)
2. Traverse the array from left to right:
    a. For each char, if it is not in the map, push it back to the current result so far.
    b. If the char is in the map, we need to decide whether use the current one or use the old one by comparing the char after the old char.
    c. If the current one makes smaller dictionary order, we delete the old one from the result and use the current char. Also need to update the map for the current position.

As for the example: ccbadbc
1. visit 'c', map empty, so result = "c", map['c'] = node(c)
2. visit second 'c', map['c'] not empty, so decide whether to use the old 'c' or current 'c', in this case, result is only one char, so doesn't matter. Let's just keep the result = "c" and map['c'] = node(c)
3. visit 'b', map['b'] empty, so result = "cb", map['b'] = node(b)
4. visit 'a', result = "cba", map['a'] = node(a)
5. visit 'd', result = "cbad", map['d'] = node(d)
6. visit 'b', now since map['b'] = node(b), compare "b" with "a", so result = "cadb", map['b'] = node(b)
7. visit 'c', since map['c'] = node(c), compare "a" with "c", so result = "adbc", map['c']= node(c)

Time O(n), space O(n) or O(1) if in-place.

ozox 发表于 2020/05/13 13:26:36
First problem:

1. Use a map, key is the char, value is a...

老哥你考虑一下 cbacbc 这个情况

if input is: cbacbc

1. first, result = cba
2. next, 'c' already in the result and char after 'c' is 'b', so use the this 'c' and delete the 'c' in the result, result = bac
3. similarly, 'b' ready in the result and char after 'b' is 'a', so use the this 'b' and delete the 'b' in the result, result = acb
4. last char is 'c', again, already exist and char after 'c' is 'b', so use this 'c' and delete the 'c' in the result, result = abc

return "abc"

新人求加米 :)

qiaofengmarco 发表于 2020-5-13 07:00
分人吧
我之前面的时候hr回复很快 描述也很清楚
可惜我碰到了不咋地的面试官 虽然水平应该很好 但不喜欢讨 ...

用bfs写的，有点长没写完

ozox 发表于 2020/05/14 07:13:52
if input is: cbacbc

1. first, result = cba
2. next, 'c' ...

How about cdacd