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  1. #1
    Propinquity is offline Member
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    Default Selection sort method?

    I'm trying to make a selection sort method that will sort a list of Strings alphabetically.
    I have a test list of Strings like so:
    Joe, Steve, Oscar, Drew, Evan, Brian, Ryan, Adam, Zachary, Jane, Doe, Susan, Fred, Billy-Bob, Cindy, Mildred, Joseph, Hammer, Hank, Dennis, Barbara

    However, whenever I run the method, the element that should go last, Zachary, in this case, ends up getting moved to the front for some reason. I'm not sure why.
    I tried changing what the first element was initialized to, to the variable i as that would logically work as well, but it ends up missing the first element in the list.
    Java Code:
    	public static void selectionStringAscendingSort (String[] words){
    		int i, j, first;
    		String temp; 
    		for ( i = 1; i < words.length; i++ )    {
    			first = 0; 						//initialize to subscript of first element
    			for(j = i; j < words.length; j ++){ 					//locate smallest element between positions 1 and i. 
    				if( words[ j ].compareTo(words[ first ]) <0 ) 
    					first = j;             
    			} 
    			temp = words[ first ]; //swap smallest found with element in position i.
    			words[ first ] = words[ i ];
    			words[ i ] = temp;     
    			System.out.println(Arrays.toString(words));
    		}
    		System.out.println(Arrays.toString(words));
    	}

  2. #2
    gimbal2 is online now Just a guy
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    Default Re: Selection sort method?

    Is this an assignment that you HAVE to do it yourself, or are you just looking to sort a set of words? Because of course the JDK provides ways to do it for you.
    "Syntactic sugar causes cancer of the semicolon." -- Alan Perlis

  3. #3
    JosAH's Avatar
    JosAH is offline Moderator
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    Default Re: Selection sort method?

    Make your outer loop iterate from 0 (not 1!) to words.length and set first to i in the top of the body of your loop.

    kind regards,

    Jos
    cenosillicaphobia: the fear for an empty beer glass

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