Insertion Sort Algorithm
Description
Insertion Sort is one of the simplest, but not very effective, sorting algorithms . It works very similar to the way (us humans) sort a hand of playing cards:
1. At first none of our cards are sorted . So we start with an "empty" space in our hand were we are going to insert cards one at a time . 
2. We take a card from our hand and we put it in the "special place" were we planned to keep our cards sorted . We do this for every card in our initial hand, until we finish them off . 
3. Finding the right position for the insertion is not hard . At this point we can apply two tactics :

In our case that “special place” were we are going to insert cards one at a time, is not an additional array, but the array itself . We know for sure that an array of size 1 is always sorted, so we consider that first sorted subarray, is the block composed by the first element .
For example, we want to sort the following array (with bold, are the elements already sorted):
8  0  3  11  5  1  14  10  1  1  2  Array is unsorted . 
8  0  3  11  5  1  14  10  1  1  2  The first sorted block is {8} . 
0  8  3  11  5  1  14  10  1  1  2  We compare {0} with {8} we move {0} at the new position, we shift {8} . 
0  3  8  11  5  1  14  10  1  1  2  We compare {3} with {8}, {0}, we move {3} at the new position, we shift {8} . 
0  3  8  11  5  1  14  10  1  1  2  We compare {11} with {8}, {11} remains at the same position . 
0  3  5  8  11  1  14  10  1  1  2  We compare {5} with {11}, {8} and {3}, we move {5} at the new position, we shift {8}, {11} . 
1  0  3  5  8  11  14  10  1  1  2  We compare {1} with {11}, {8}, ..., {0}, we move {1} at the new position, we shift {0}, {3}, ... {11} . 
1  0  3  5  8  11  14  10  1  1  2  We compare {14} with {11}, we move {11} at the new position . 
1  0  3  5  8  10  11  14  1  1  2  We compare {10} with {14}, {11}, {8}, we move {10} at the new position, we shift {11}, {14}. 
1  0  1  3  5  8  10  11  14  1  2  We compare {1} with {14}, {11}, ..., {0}, we move {1} at the new position, we shift {3}, {5}, ..., {14} . 
1  0  1  1  3  5  8  10  11  14  2  We compare {1} with {14}, {11}, ..., {1}, we move {1} at the new position, we shift {3}, {5}, ..., {14} . 
2  1  0  1  1  3  5  8  10  11  14  We compare {2} with {14}, {11}, ..., {1}, we move {2} at the new position, we shift {1}, {0}, ..., {14} . 
The pseudocode for the algorithm can be easily written as follows:
FUNCTION SORT(A, length)
// The array is 0indexed
FOR i:=1 TO length DO
key := A[i]
j := i  1
// COMPARE
WHILE j >= 0 AND A[j] > key DO
// SHIFT
A[j+1] := A[j]
j := j  1
// MOVE
A[j+1] = key
Java Implementation
The Java implementation of the algorithm is pretty straightforward:
public class InsertionSort {
// Print array
public static void printArray(int [] array){
for(int i : array){
System.out.printf("%d ", i);
}
}
// Sort array through the 'Insertion Sort' method .
public static void sortArray(int [] array){
int key, j;
for(int i = 1; i < array.length; ++i){
key = array[i];
j = i  1;
while(j >= 0 && array[j] > key){
// Shift array elements so we can place
// the key to the right place
array[j+1] = array[j];
j;
}
// Place the key into position
array[j+1] = key;
}
}
public static void main(String[] args){
int [] array = new int[] {8, 0, 3, 11, 5, 1 , 14, 10, 1, 1, 2};
sortArray(array);
printArray(array);
}
}
Comments