O(N) solution to find Nth ugly number.

ugly_number.py

def ugly_number(n):
    """ Returns Nth ugly number """

    assert n > 0, 'The argument must be greater than zero.'
    
    list = [0]*n
    i2, i3, i5 = 0, 0, 0
    w2, w3, w5 = 2, 3, 5
    w, list[0] = 1, 1
    
    for i in range(1, n):
        w = min(w2, w3, w5)
        list[i] = w

        if w == w2:
            i2 = i2+1
            w2 = list[i2]*2

        if w == w3:
            i3 = i3+1
            w3 = list[i3]*3

        if w == w5:
            i5 = i5+1
            w5 = list[i5]*5

    return w


if __name__ == "__main__":
    print 'ugly numbers'
    print 'first 20:',
    for n in range(1, 21, 1):
        print ugly_number(n),

    print ''
    print '10th:',
    print ugly_number(10)
    print '100th:',
    print ugly_number(100)
    print '300th:',
    print ugly_number(300)
    print '600th:',
    print ugly_number(600)
    print '1200th:',
    print ugly_number(1200)
    print '2400th:',
    print ugly_number(2400)
    print '4800th:',
    print ugly_number(4800)
    print '9600th:',
    print ugly_number(9600)
    print '50000th:',
    print ugly_number(50000)