QUOTE(Martyr2 @ 21 Aug, 2007 - 11:53 AM)
The solution I have designed for you have a bit of "kookyness" in it so I will have to explain what I am doing and why. I created this solution using an actual selection sort algorithm because I wasn't sure if you wanted the selection sort you mention in the topic or the bubble sort you provided. So I wrote it with the idea that you could drop in a bubblesort if you needed by replacing the function and making sure it takes an array and returns a sorted array.
Now I am not a big fan of changing a person's program all around more than need be to get a solution. In sticking with this mentality, I tried to build on what you had implemented already like using charAt() and primitive datatype arrays etc. One place where this gets sticky is when converting a charAt over to the integer value needed for sorting appropriately. I pass the char to the valueOf to get it to a string that I could then use in a parseInt to make sure it was an integer. The solution works out nicely, but it looks funny doing it.
So here is the solution...
CODE
import java.io.*;
import java.util.*;
public class String1 {
public static Scanner con = new Scanner(System.in);
// Main program start point.
public static void main(String [] args) {
int length;
System.out.println("Input a string of letters and numbers:");
String str = con.next();
length = str.length();
process(str,length);
}
// Process the entered input and determine if it is a number.
public static void process (String str1, int length1)
{
int i = 0, numbercount = 0;
int[] numbers = new int[length1];
// Initialize entire array with invalid value
for(int f = 0; f < numbers.length; f++) {
numbers[f] = -1;
}
for (i=0; i<length1; i++)
{
// If number, convert it to an integer to store
if(Character.isDigit(str1.charAt(i))) {
numbers[numbercount] = Integer.parseInt(String.valueOf(str1.charAt(i)));
numbercount++;
}
}
// Sort the array and store it in new sorted array
int[] sortednumbers;
sortednumbers = selectionSort(numbers);
// Now print it
printArray(sortednumbers);
System.out.println();
}
// This is a custom selection sort, but you could easily make it bubble
// by making sure it takes an integer array and returns a sorted array.
private static int[] selectionSort(int[] numArray) {
for (int i = 0; i < numArray.length - 1; i++) {
for (int j = i + 1; j < numArray.length; j++) {
if (numArray[i] > numArray[j]) {
int swap = numArray[i];
numArray[i] = numArray[j];
numArray[j] = swap;
}
}
}
return numArray;
}
// This function loops through our sorted numbers and prints
// only those elements which are numbers
private static void printArray(int[] printArray) {
String result = "";
for (int i = 0; i < printArray.length; i++) {
if (printArray[i] > -1) {
if (result.equals("")) {
result += printArray[i];
}
else { result += ", " + printArray[i]; }
}
}
System.out.println("Sorted Numbers: " + result);
}
}
One thing I want to point out is the use of the -1 value in the array to mark invalid array entries. I look for this in the print to know if I have actually have a legit value in the array since the selection sort will push all legit values to the end of the array. I could reverse the sort order in the algorithm, but not knowing if you are going to change it later I didn't bother.
To make a more elegant solution I would have considered using an ArrayList or implement a Linked List so that I didn't have to loop through the entire array looking for valid entries. I again wasn't sure how far along you were and if you had experience in those solutions. So I opted for the more basic way.
Hope you find this solution ok since it does what you want, modify it whatever way you like.
excellent solution, however I find checking values (case select style) much easier. or you could to a if statement to pull out the numbers from the string.
i know this is just psudocode, but i'm not writing it for you.
CODE
if the substring(i) of str = 1 then put in strNum
else if the substring(i) of str = 2 then put in strNum
.
.
.
else if the substring(i) of str = 0 then put in strNum
this would be in a for loop that increases i each time until it finishes str
in other words
CODE
for(int i == 0; i <= str.length(); i++)
{
}
you could also shorten the if statement by putting or's, but it may go off screen.
CODE
if the substring(i) of str = 1 or the substring(i) of str = 2 or...the substring(i) of str = 0
then put in strNum
then run the sort and you should be good to go.
that's alot of writing! good luck!
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