检查递归中的2个字符

Question

我想检查这个“a”和“b”是否在userInput中。 例如：用户要输入“hello”，它会说它不是ab字。 例如：如果用户输入“bag”，它会说它是一个ab字。 例如：如果用户输入“年龄”，则会说它不是ab字（因为缺少b）。

我已经尝试了很多方法，包括使用计数器来计算字符串在字符串内有多少次。 如果“a”和“b”位于字符串内部，它将打印“内部”，否则如果其中一个不在里面或两个都不在里面，它会打印“不在里面”

请帮助我出。我试过了，但似乎无法正常工作。

def tryword(x):   
    if x == '': 
     return 0 
    lastNum= x[-1] 
    if (lastNum == "a" and "b"): 
     return tryword(x[:-1]) + 1 
    else: 
     return tryword(x[:-1]) + 0 

if recursiveWord >= 0: 
    print "inside" 
else: 
    print" not inside" 

print tryword("abcdef") 

Answer 1

您需要添加两个累加器参数来跟踪您是否找到每个字符。每个布尔值为一个布尔值。

def trywordp(x, foundA, foundB): 
    if foundA and foundB: return True # Positive base case: Both found! 
    if x == "": return False # Negative base case: String exhausted 
    if x[0] == 'a': return trywordp(x[1:], True, foundB) # First character in string is 'a', continue checking next character 
    if x[0] == 'b': return trywordp(x[1:], foundA, True) # Ditto for 'b' 
    return trywordp(x[1:], foundA, foundB) # Neither 'a' nor 'b', continue looking 

def tryword(x): 
    return trywordp(x, False, False) # We have not found any of them yet so start with foundA and foundB as False 

if tryword("stringwithaandb"): 
    print "inside" 
else: 
    print "not inside" 

Answer 2

检查你的病情

if (lastNum == "a" and "b"): 

您大概意思是这样

if (lastNum == "a" or lastNum == "b"): 

但这不会工作。你可以使用魔术数字。

def tryword(x):   
    if x == '': 
     return 0 
    lastNum= x[-1] 
    if (lastNum == "a"): 
     return tryword(x[:-1]) | 1 
    if (lastNum == "b"): 
     return tryword(x[:-1]) | 2 
    return tryword(x[:-1]) 

assert tryword("qwer") == 0 
assert tryword("a") == 1 
assert tryword("b") == 2 
assert tryword("bb") == 2 
assert tryword("ab") == 3 
assert tryword("abababab") == 3 

def check_inside(s): 
     if tryword(s) == 3: 
      print "inside" 
     else: 
      print "not inside"