Find Duplicate in Linear Time

You are given an array of numbers that contains all numbers from 0 to n, in some random order, except that one number is missing and another number is repeated (thus, there are still n numbers in the array). Find the repeated number in linear time and do not use any other data structure.

The numbers in the array can just be placed in the appropriate place in the array, just use the number as it's own position index. If when you go to place a number i in position i you notice that a[i] already contains i then you know you found the duplicate. If not, place it and use the number that was there as your new index.

#!/usr/bin/python
from Numeric import *

def find(a):
    i = 0
    while True:
        print a
        if (a[i] == i):
            i += 1
            continue
        if (a[a[i]] == a[i]):
            print "Repeat is %d" % a[i]
            return a[i]
        c = a[a[i]]
        a[a[i]] = a[i]
        a[i] = c
        i = c

        
#Test cases
a = array([0,3,2,1,4,4,6])
a = array([0,1,2,3,4,4,6])
a = array([4,4,2,3,0,1,6])
a = array([4,4,0,1,2,3,6])

find(a)