Selection Sort using Groovy

Question

Here's my code for performing a selection sort using Groovy:

class SelectionSorting {
    void sorting() {
        def sortmethod = {
            List data = [ 1, 5, 2, 3, 7, 4, 6, 8, 9, ]
            def n = data.size()
            println "Before sort : " + data
            for(def i in 0..n) {
                 def position=i
                 for(def j in i+1..n) {
                    if(data[position] > data[i])
                        position=i
                 }
                 if(position!=i) {
                    swap(data[i],data[position])
                 }
            }
            println "After sort : " + data
       }
       sortmethod()
    }
}

SelectionSorting s = new SelectionSorting()
s.sorting()

However, the output I see is still an unsorted array:

Before sort : [1, 5, 2, 3, 7, 4, 6, 8, 9]
After sort : [1, 5, 2, 3, 7, 4, 6, 8, 9]

I am very new to Groovy. I'm supposed to insert the logic within a closure only. I'm not sure what I need to change in the closure I have created in my code above. Please help.

Answer 1

You are using the wrong index when calculating the position of the minimum value; you should use j instead of i (added println to show iterations):

def selectionSort = { data ->
    int n = data.size()
    for (int i = 0; i < n - 1; i++) {
        // Find the index (position) of the minimum value
        position = i       
        for(int j = i + 1; j < n; j++) {
            if(data[j] < data[position]) {
                position = j               
            }
        }

        // Swap
        if (position != i) {
            temp = data[position]
            data[position] = data[i]
            data[i] = temp
        }

        println data
    }
    println "result: $data"
}

So that selectionSort([1,5,2,4,3,8,7,9]) yields:

[1, 5, 2, 4, 3, 8, 7, 9]
[1, 2, 5, 4, 3, 8, 7, 9]
[1, 2, 3, 4, 5, 8, 7, 9]
[1, 2, 3, 4, 5, 8, 7, 9]
[1, 2, 3, 4, 5, 8, 7, 9]
[1, 2, 3, 4, 5, 7, 8, 9]
[1, 2, 3, 4, 5, 7, 8, 9]
result: [1, 2, 3, 4, 5, 7, 8, 9]