Java Decode Machine Questions

/**
*
* Java writing Encode method and Decode method, machine questions please use java,c,c++
* In any of the languages implement two functions encode () and Decode (), respectively, to implement the transformation and restoration of the string.
* Transform function encode () order to see the characters that know the string, and generate a new string per group according to the following rules:
* (1) If the current character of the string is not known to be a numeric character greater than 0, the character is copied with the new string;
* (2) If the current character of a known string is a numeric character, and he does not have subsequent characters, simply copy it to the new string;
* (3) If the current character of a known string is a numeric character greater than 0, and there are subsequent characters, set the value of the numeric character to N,
* The subsequent character (including the subsequent character is a numeric character) is duplicated n+1 times to the new string; (4) In a group of the above transformations, insert an underscore ' _ ' between different groups for separation;
* (5) If the string contains an underscore ' _ ', then the transform is to use "/ul". For example: the Encode () function transforms the string 24ab_2t2 to
* 444_aaaaa_a_b_/ul_ttt_t_2
*
*
*/
public class Teststringencodedemo {

public static String pub = "";

public static void decode (String str) {

/*
* First operation judgment ' _ ', all subsequent operations are the string after recursion
*/
if (Str.charat (0) = = ' _ ') {
Pub = pub + "/ul" + "_";
} else if ("123456789". IndexOf (Str.charat (0)) = =-1) {//To determine if the character is a numeric type
Pub + = Str.charat (0) + "_";
} else if (str.length () = = 1) {//If the string has only one pop-out method
Pub + = str;
Return
} else {
/*
* "123456789". IndexOf (Str.charat (0)) +1
* Using this method to get the word Poute is the value of the character (since it is all plus 1 from 0 bits) the need to go is the literal value plus 1 all direct +2
*/
for (int i = 0; i < "123456789". IndexOf (Str.charat (0)) + 2; i++) {
Pub + = Str.charat (1);//takes the latter of the current string
}
Pub = pub + "_";
}

Recursively intercepts a string (instead of looping through a string and extracting a value that determines whether the character is an int)
if (Str.length ()! = 1) {
Teststringencodedemo.decode (str.substring (1));
}
}

public static void Encode (String str) {
String pub = "";

for (int i = 0; i < str.length (); i++) {
if (Str.charat (i) = = ' _ ') {//If there is an underscore ' _ ' in the known string, the transform is "/ul"
Pub + = "/ul";
} else if ("123456789". IndexOf (Str.charat (i), 0) = =-1) {//The current character of the string is not a numeric character greater than 0, copy the character with the new string
Pub + = Str.charat (i);
Stitching the last one
} else if ("0123456789". IndexOf (Str.charat (i), 0)! =-1 && i = = (Str.length ()-1)) {
Pub + = Str.charat (i);
}
The current character of the string is a numeric character greater than 0, and there are subsequent characters,
else if ("0123456789". IndexOf (Str.charat (i), 0)! =-1 && i! = Str.length ()-1) {
int pool = Integer.parseint (Str.charat (i) + "");

for (int j = 0; J <= Pool; j + +) {
Pub + = Str.charat (i + 1);
}
}
Pub + = "_";
}
Pub = pub.substring (0, Pub.length ()-1);
SYSTEM.OUT.PRINTLN (pub);
}

public static void Main (string[] args) {
char c = "12345678". CharAt (0);
System.out.println (Character.getnumericvalue (c) + 2);
System.out.println ("12345678". CharAt (0));
String str = "24AB_2TT";

Decode (str);
SYSTEM.OUT.PRINTLN (pub);
Encode (str);
}

}

Java Decode Machine Questions