[第二轮] 2/18-2/24 CareerCup 1.1

czhoume 发表于 2013-2-18 06:03
I think no more data structure allowed need to be more specified. I will ask if I am in an interview ...

Nice job, I would do the same way if I hadn't seen the solution..

1. Hashtable
2. Sort & Check

https://gist.github.com/JasonGitHub/4975301

paradox 发表于 2013-2-18 03:12
I think you need improve your code style. Your python code looks very messy.
Here is mine, you ca ...

For the ifDuplicate1, you can check the value every time before adding 1 and return false if it is already 1. I think there is no need to write another for loop to check the charCount values.

By the way, I don't see what is the additional data structure in ifDuplicate1.

The code is neat. ;)

Language: C
Algorithm: Nested loop, compare the i-th with the rest

https://gist.github.com/liangkaile/4975514

gist link:

Java solutions: https://github.com/wangtz/Career ... dStrings/Ch1Q1.java

n: input.length
k: amount of different unique characters

for time and storage complexity:
* hash o(n) o(k)
* simple hash only works for 'a' to 'z' can be extended o(n) o(k)
* bit operation only works for 'a' to 'z' can be extended to UNICODE using multiple ints o(n) o(k/64)
* sort before check o(nlogn) o(1)
* brute force nested loop compare o(n^2) o(1)

https://gist.github.com/anonymous/4976549
因为以前做过careercup的一部分题，所以思路大概是差不多的！但是难免有错漏和效率方面的问题，请多多指教！

https://gist.github.com/JerrySun363/0c7fb3b4d9c54aa41171

其实不是很清楚能用什么样的API
第一个用一个HashSet，第二个用了排序+比较

用了最笨的办法，每个字符循环比较。时间，空间复杂度什么的，还不知道怎么算，明天学习一下，然后看一下大家的做法
https://gist.github.com/sunxw/4975738

yukainuo23 发表于 2013-2-18 18:45
https://gist.github.com/anonymous/4976549
因为以前做过careercup的一部分题，所以思路大概是差不多的！ ...

看了一会才明白，ls是把 unique characters 当作输入并且放到bitmap里去的；
大概对题目的理解不一样吧... ...