问题描述
分配:
编写一个名为is_subsequence的递归函数,该函数需要两个 字符串参数,如果第一个字符串是 第二个字符串的子序列,但否则返回False。我们说 如果您可以通过以下方式得出A,则字符串A是字符串B的子序列 从B删除零个或多个字母,而不更改其顺序 剩下的字母。您可以假设两个字符串都不包含 大写字母。
到目前为止,我所拥有的(必须使用此格式):
def is_subsequence(strA,strB = None):
if strB == None:
strB = []
print(is_subsequence("dog","dodger"))
我的问题:
我无法弄清楚如何通过删除字母以赋值方式递归地编辑字符串。 (strB.replace("e","")
不起作用。
解决方法
根据指示,我不知道是否允许这样做,但是我添加了索引参数,并且能够提出这个建议。 (代码段是Python)
def is_subsequence(strA,strB,index=0):
"""
Function that determines if string A is a subsequence of string B
:param strA: String to compare subsets to (strings are immutable)
:param strB: Parent string to make subsets of (strings are immutable)
:param index: number of combinations tested
:return: True if a subsequence is found,False if index reaches maximum with none found
"""
binary = '{:0{leng}b}'.format(index,leng=len(strB))
if index > 2**len(strB)-1: #If past maximum return False
return False
def makesub(sub="",pos=0):
"""
Function that builds up the current subsequence being tested by using binary to choose letters
:param sub: The currently selected letters
:param pos: Position in binary string to compare
:return: completed subsequence for comparing to strA
"""
if pos > len(binary)-1: #If pos is passed the length of our binary number return the created subsequence
return sub
if binary[pos] == "1": #If the pos in binary is a 1,add the letter to our sub. If it is 0 do nothing.
sub = sub + strB[pos]
return makesub(sub,pos + 1) #increment pos and try again or return what the call returned
if makesub() == strA: #If the returned sub matches strA we found a matching subsequence,return True
return True
return is_subsequence(strA,index + 1) #increment index and try again or return what the call returned
print(is_subsequence("dog","dodger"))