usage of the append function of list in python

Question

class Solution:    
    # @param num, a list of integer
    # @return a list of lists of integers
        def permute(self, num):
            self.res = [];
            self.dfs(num, 0)
            return self.res

        def dfs(self, num, level):
            if level == len(num):
                self.res.append(num)
                print(num)
                return
            for i in range(level, len(num)):
                num[i], num[level] = num[level], num[i]
                self.dfs(num, level+1)
                num[i], num[level] = num[level], num[i]

The above code is used to generate all permutations given a collection of numbers. For example, num = [1, 3] The result will be: [1 3], [3, 1]

But there is a bug for the above code that I don't understand, which is self.res.append(num) . If I change it to self.res.append(num[:]) , then the code is correct. Can anyone explain why?

Using self.res.append(num) , the result is:

[1, 3], [1, 3]

Using self.res.append(num[:]) , the result is:

[1, 3], [3, 1]

Answer 1

The elements of a python list are references to other objects. When you append using self.res.append(num) , the list is grown by 1 element, and the last element is set to refer to the object pointed to by num .

Now in the first case, there are 2 references to the same list object. As self.res[0] and self.res[1] refer to the same object, all changes performed through either are also visible through the other.

In the second case, using num[:] , [:] operator makes a new list that is copy of the original.

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For general algorithm for creating all permutations of a given collection of elements, use the itertools.permutations :

>>> from itertools import permutations
>>> print(list(permutations([1, 3])))
[(1, 3), (3, 1)]