Lintcode Easy title: Space Replacement Space replacement

Topic:

Space Substitution

Design a method that replaces all the spaces in a string with a %20 . You can assume that the string has enough space to add new characters, and you get the "real" character length.

Sample Example

For strings "Mr John Smith" , the length is13

The result after replacing a space is"Mr%20John%20Smith"

Note

If you are using Java or Python, use a character array in your program to represent the string.

Solving:

A problem that indicates the lowest pass rate ....

Link it with a Java string, then ToCharArray, then go out and pick up the same as before, the original%20 represents a space .... Reference program, directly in the original Char array operation, the first new string array of length, and then insert the% 20, so that the solution, for the post-insertion solution encountered, the topic will give such a hint: assuming that the string has enough space to add new characters , according to the question to find the answer , look for ideas.

Java Program:

 Public classSolution {/**     * @paramString:an Array of Char *@paramlength:the True Length of the string *@return: The true length of new string*/     Public intReplaceblank (Char[] String,intlength) {        //Write Your code here        intReallen =length;  for(inti = 0;i<length;i++){            if(String[i] = = ") Reallen+ = 2; }        intindex =Reallen;  for(inti = length-1;i>= 0; i-- ){            if(String[i] = = ") {string[--index] = ' 0 '; string[--index] = ' 2 '; string[--index] = '% '; }Else{string[--index] =String[i]; }        }                returnReallen; }}

View Code

Total time: 1137 Ms

Python program:

classSolution:#@param {char[]} string:an array of char    #@param {int} length:the True length of the string    #@return {int} The true length of new string    defReplaceblank (self, string, length):#Write Your code here        ifstring==None:returnNone Reallen=length forSiinchstring:ifsi==' ': Reallen+ = 2Index=Reallen forIinchRange (length-1,-1,-1):            ifString[i] = =' ': Index-= 1string[Index]='0'Index-= 1string[Index]='2'Index-= 1string[Index]='%'             Else: Index-= 1string[Index]=String[i]returnReallen

View Code

Total time: 393 Ms

Lintcode Easy title: Space Replacement Space replacement