Selection sort: storing value instead of index

Question

I'm studying sorting algorithms, including selection sort, so i decided to write a method and it works fine, but when i checked the book it had 2 variables so i checked it and found that it's using a variable to store the current index and the other as temporary to swap

while mine had only the temporary variable that also stored the initial value in the index as the lowest, then compared it to the other values in the array and swapped if a larger value was found.

Here's my code:

public static void selectionSort(int[] arr){
int lowest;
for(int i = 0; i < arr.length - 1; i++){
  lowest = arr[i];
  for(int j = i+1; j<arr.length; j++){
    if(arr[j]<lowest){
      lowest = arr[j];
      arr[j] = arr[i];
      arr[i] = lowest;
    }
   } 
  }
 }

and Here's the book's

public static void selectionSort(int[] list){   
    int min; 
    int temp;
    for(int i = 0; i < list.length - 1; i++) { 
        min = i;
        for(int j = i + 1; j < list.length; j++)
            if( list[j] < list[min] )
                min = j;
        temp = list[i];
        list[i] = list[min];
        list[min] = temp;
    }
}

so I looked on the web and all of them follow the same way as the book, so is my code bad or slower, or it is not considered Selection sort ?

sorry for any english mistakes :P

Answer 1

So, the original code works like this:

Start with the first element of the array
Find the smallest number in the array
Swap the two numbers
Repeat the process Until You reach the end of the list

While Yours does this:

Start with the first element of the array
For each element smaller than the current Swap the two numbers
Replace the lowest number with the swapped
Repeat the process

The result should be the same, but probably You are swapping more numbers than the first one. This probably will make it a little bit slower than the original.

Actually it kinda looks like the insertion sort now, so basically You are swapping the elements with all that are bigger than the one You have.