Backtracking - full array

= A [, 2, 3 ] 
the n- = len (A) 
RES = []
 DEF Permutation (A, Solution):
     # Note that this use global modification res, in order to update the results of 
    global RES
     IF len (A) == 0 : 
        res.append (Solution) 
        return 
    for I in Range (len (A)): 
　　　　　newsolution = + Solution [A [I]] 
        new _a = A [: I] + A [I +. 1 :] 
        Permutation (new _a, newsolution) 
Permutation (A, []) 
Print (RES) 
output: 
[[ 1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

The basic idea:

In fact, for backtracking, we need to start thinking about the reverse. Every time we choose from the original array was added to a result, when there is no element in the original array (new) (i.e. len (a) == 0, this time the result is [1,2,3]), we to give the first arrangement, the arrangement will be added to the result set, and then return to the previous step, that is, we have now [1,2], and then return to step [1], at which time was added 3, then add 2 to give [1,3,2],

And so on.