[第二轮] 2/18-2/24 CareerCup 1.3

先排序呀，再判断
Arrays.sort -->  Arrays.toString
https://gist.github.com/lazyGoose/4988562

另外，哪位能教教我怎么加链接...

先判断是否长度相等，再对两个string排序，最后判断sorted string是否相等

https://gist.github.com/skxie/4b4a5eb6ac2cdef047a2

https://github.com/quantumrose/C ... .3.Permutation.java

method1: quicksort two strings and compare
methdo2: compare character counts

新手学习
思路：
     1、sort分别排序后，从第一个字符开始依次比较两个string字符是否相等。
     2、使用bitmap储存字符出现次数。两个string的相同字符的数目必须相等。


https://gist.github.com/KuangyinWang/4992851

lazyGoose 发表于 2013-2-20 02:34
先排序呀，再判断
Arrays.sort -->  Arrays.toString
https://gist.github.com/lazyGoose/4988562

同求，我也没加上。

https://gist.github.com/KainuoYU/4993144

这里有两种方法，一种是sort再比较，效率是O(nlgn），另外一种是用数组数不同字符个数，思想应该和hashset一致，效率为O(n）

吐下糟，由于string，stringbuffer，char[]的基本java操作不熟，在这题上浪费了太多的时间。第一次感到C是真方便。。。。


https://github.com/1094401996/Ca ... dot3/RemoveDul.java

Java language
https://gist.github.com/mettaworld/4995136
make a little complex algorithm. It processes string in this way: abcdabcd - a2b2c2d2
Then compare the first string with the second one.

Assume they are ascii string
1 sort
2 compare
https://gist.github.com/sunxw/4995674

compare the length of two strings; if they are equal then create boolean array of size 256, initialized false
go through original string, set corresponding one in the array true when finding each character
go through the second string, check the boolean array when finding each character
https://github.com/swang1022/car ... 3/isPermutation.cpp