[英]Iterative solution for :- Finding String permutations
I read this simple and elegant python solution for finding all permutations of a given string. 我阅读了这个简单而优雅的python解决方案,用于查找给定字符串的所有排列。 It is recursive.
它是递归的。 Based on that I tried to implement an iterative solution in python.
基于此,我尝试在python中实现迭代解决方案。
Below is my code. 以下是我的代码。 But it works only for 3 character strings :( Stuck trying to see how the recursion base case condition and recursion condition translates into iterative(non-recursive) Any pointers would help to get a iterative solution working.(Either based on this algorithm or any other)
但它仅适用于3个字符串:(试图看看递归基本情况条件和递归条件如何转换为迭代(非递归)任何指针都有助于使迭代解决方案工作。(基于此算法或任何其他)
def permutations_iter(word):
while True:
perms = []
result = []
char = word[0]
new_word = word[1:]
if len(new_word)==2:
perms = [new_word,''.join(reversed(new_word))]
for perm in perms:
#insert the character into every possible location
for i in range(len(perm)+1):
result.append(perm[:i] + char + perm[i:])
return result
if len(new_word)==2:
break;
#example code to call this iterative function
print permutations_iter("LSE")
You can convert every recursion to an iteration using a stack. 您可以使用堆栈将每个递归转换为迭代。 But in this case it's even simpler since the algorithm is very simple.
但在这种情况下,由于算法非常简单,因此更简单。
def perms(word):
stack = list(word)
results = [stack.pop()]
while len(stack) != 0:
c = stack.pop()
new_results = []
for w in results:
for i in range(len(w)+1):
new_results.append(w[:i] + c + w[i:])
results = new_results
return results
For a more general conversion of recursion to iteration with a stack read this 对于递归到迭代的更一般转换,使用堆栈读取此内容
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