insertion sort does not order array correctly

Question

Here is my insertion sort, exactly the way it is in the book "introduction to algorithms":

def insertion_sort():
    A = [5,2,4,6,1,3]
    for j in range(1, len(A)):
        print 'j:'+str(j)
        key = A[j]
        print 'key:'+str(key)
        i=j-1
        print 'i:'+str(i)
        while i > 0 and A[i] > key:
            A[i+1] = A[i]
            i=i-1
            print 'new i: '+str(i)
        print 'swapping value: '+str(A[i]) + ' with value: '+str(A[i+1])
        print ' '
        A[i+1] = key
    print A

This prints:

[5, 1, 2, 3, 4, 6]

What am I doing wrong to have them out of order?

C.B. · Accepted Answer · 2014-03-11 17:59:19Z

5

In Introduction to Algorithms, they always assume arrays start at index 1, so you are starting your range() at 1, but python lists are 0-based indexed. This means you are never comparing 5, which is at A[0]. Notice everything after 5 is sorted.

modifying your for loop to -

for j in range(0, len(A)):

and your while condition to

while i >= 0 and A[i] > key:

Should do the trick.

answered Mar 11, 2014 at 17:59

C.B.

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