[第二轮] 2/18-2/24 CareerCup 1.4

Kimurate 发表于 2013-2-20 10:43
Just copy the char array to a new array and make some change in process.
C language: https://gist.g ...

Your code doesn't work.
There is a compiling error reported by mingw(gcc 3.4.5), in line 40, you should change your code into:
char *replaceStr = (char*) malloc(sizeof(char)*(strlen(str)+countSpace*2));

And output is not correct, either....you should add:
replaceStr[j] = '\0';
at the final stage of your code.

moophis 发表于 2013-2-22 10:55
Your code doesn't work.
There is a compiling error reported by mingw(gcc 3.4.5), in line 40, you  ...

Thx for advice.
The code worked well on my machine, it should be a problem of the gcc version, which is 4.7 or llvm-gcc-4.2.
I modify my code as your advice, thx again.

先找到从最右边开始第一个不是空格的位置i，然后再用一个变量j从最后一个位置开始扫，
如果str是空格，str[j-2..j]替换成"%20"
直到i =j，说明空格填满

https://gist.github.com/skxie/b9d8d368eaa1b6fbdf93

https://gist.github.com/ThunderXu/5018072
先算出替换空格后string的结束位置，再从此结束位置开始由后向前的进行再填充，中途遇到空格则依次填充'0''2''%'

https://gist.github.com/bigteeth/5021750

算法不知道为什么  后面多存储了一些字符  

It is kind of late

overveiw'

1 calculate new lengh
2 replace spaces with %20
https://gist.github.com/sunxw/5022902

1. count the total number of space in the string.
2. subtract the spaces at the beginning and at the end.
3. create a new string to store output string
4. copy output string to input string

https://gist.github.com/Andy1989/69f32cc3ea90c34e7cd8

https://gist.github.com/yzl232/5035524

用python。 还简单。

https://gist.github.com/flatlight/5056328
first calculate the length, then replace backwards in order to replace in place

yzl232 发表于 2013-2-26 11:07
https://gist.github.com/yzl232/5035524

用python。 还简单。

原题的意思应该是做in place的替换..即空间复杂度是O(1)才对..