[第二轮] 2/18-2/24 CareerCup 1.4

Write a method to replace all spaces in a string with '%20'. You may assume that the string has sufficient space at the end of the string to hold the additional characters, and that you are given the "true" length of the string. (Note: if implementing in Java, please use a character array so you can perform this operation in place.)

EXAMPLE
Input  : "Mr John Smith    "
Output: "Mr%20John%20Smith"

发帖规范:
http://www.1point3acres.com/bbs/thread-48094-1-1.html
http://www.1point3acres.com/bbs/thread-32423-1-1.html

overview:
1. Create a new empty string
2. Go through the given string character by character,
     i.  If there is an non-space character, then just add it to the new string
     ii. If there is an space character, then add "%20" to the new string instead
3. Return the new string when loop ends

https://gist.github.com/edussx/4970604

edussx 发表于 2013-2-17 15:56
overview:
1. Create a new empty string
2. Go through the given string character by character,

他的意思应该是 in place?

所以
1. Read the string and compute the new length of this string if we replace it
2.
for (int i=newLen,j=oldLen;j>=0;--j){
  if (str[j]==' ') str[i--]='0',str[i--]='2',str[i--]='%';
  else str[i--]=str[j]
}

这样大概就可以了吧..

假设string长度为n, str[1]到str[m]为原来的字符，从str[n-1]开始向前扫描，到发现str[m]为止，distance = n-1-m就是str[m]要往后移动的距离
往前继续扫描的过程中，假设str[j]是空格，就把“%20”插入到str[j+distance]的为止，由于多插入了两个字符，所以distance = distance-2，以此类推..

https://gist.github.com/chrislukkk/4971010

create another array with new len which is equal to length of original input + 2*space number. copy characters from end of array to the new array.

https://gist.github.com/njcongtou/4971371

Step 1: count the length of new string， length_new_string = strlen(str) + num_space * 2;
Step 2: replace all spaces in origin string by two cursors. if str[j] equals ' ',  str[i, i-1, i-2] are replaced by “%20” else str[i] is replaced by origin element.
https://github.com/Tedatworking/ ... ster/20130217_4.cpp

jimwallet 发表于 2013-2-17 20:53
create another array with new len which is equal to length of original input + 2*space number. copy  ...

写的挺清晰的，就是我感觉在原来的string上就能转换。

Solution 1: implement without in place by using build in String func
Solution 2: in place.
       o examine the string backward.
       o if not space, just copy to back.
       o else insert '0' '2' '%'
https://gist.github.com/mazl123321/4973151

https://gist.github.com/cqx83/4973856
先扫一遍String，数空格数量，定义char[]的长度，然后逐个扫入char[]中，遇到空格变为%20.
O(n)

1. Sort & Check (Time: n* log(n), Space: log(n))
2. Hashtable (Time: n, Space: 256)

https://gist.github.com/JasonGitHub/4975606