擅长:python、mysql、java
<p>实际上,itertools已经为您解决了这个问题。</p>
<pre><code>import itertools
allp = [x for x in itertools.permutations(range(3))]
print allp
mylist = ['A','B','C']
allp2 = [x for x in itertools.permutations(mylist)]
print allp2
</code></pre>
<p>输出</p>
<pre><code>[(0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0)]
[('A', 'B', 'C'), ('A', 'C', 'B'), ('B', 'A', 'C'), ('B', 'C', 'A'), ('C', 'A', 'B'), ('C', 'B', 'A')]
</code></pre>