使用重复元素进行全置换的非递归算法? [英] Non-recursive algorithm for full permutation with repetitive elements?

问题描述

我通常在Python中使用以下递归算法:

I usually use the recursive algorithm as follows in Python:

def permutate(array, t, n):             

    if t == n:
       for i in range(n):
            print array[i]

       return

   for j in range(t,n):
      flag = 1
      for r in range(t,j):
          if array[r] == array[j]:
              flag = 0
              break

      if flag == 0:
          continue
      else:
          array[j],array[t] = array[t],array[j]
          permutate(array,t+1,n)
          array[j],array[t] = array[t],array[j]

这很整齐. 但是我希望找到一种方便的，非递归的算法来对重复元素进行完全置换吗?

This one is neat. But I hope to find a convenient, non-recursive algorithm to do full permutation with repetitive elements?

推荐答案

这是取消递归"递归函数的一种通用方法:假设我们具有以下递归函数:

Here is a generic way to "un-recursify" a recursive function : Let's say we have the following recursive function :

RecFunc (parameters)
    ...
    ...
    var x = RecFunc (other parameters)
    ...
    ...
EndFunc

要取消递归"，可以使用如下所示的堆栈:

To "un-recursify" it, you can use a stack like this :

NonRecFunc (parameters)
    stack MyStack;
    MyStack.push (InitialValue);
    While (MyStack is non empty)
       var S = MyStack.pop;
       # begin operations with S
       ....
       # results are x_1, ..., x_n
       for x_i = x_1 to x_n
           MyStack.push (x_i);
       endfor
   endWhile
EndFunc

在您的情况下，堆栈包含一个由数组和一个int组成的对.初始值为输入中的数组，并且int m = 0.操作看起来像这样

In your case, the stack contains a pair consisting of an array and an int. The initial value is the array in input and the int m=0. The operations could look like this

for i = m to n
    for j = i+1 to n
       if array[i] == array[j]
           continue
       endif
       array_c = copy of array
       permute entries i and j in array_c
       push (array_c, m+1) in the stack
    endfor
endfor

祝你好运！

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