我需要编写一个对句子中的字符进行线性搜索的程序。除了print()
之外,我必须不使用任何内置函数。
程序应输出字符所在的索引。
如果字符不在句子中,则应输出-1作为索引。
我的代码输出以下内容:
Enter a sentence: Hello
Enter a character: e
The character is -1.
The character is 1 on the index.
即使只输出:
Enter a sentence: Hello
Enter a character: e
The character is 1 on the index.
下面是我的代码:
def linear_search(intList,target):
found = False
count = 0
while count < len(intList):
if intList[count] == target:
print("The character is", count, "on the index.")
found = True
break
if intList[count] != target:
print("The character is -1.")
count = count + 1
return count
sentence = input('Enter a sentence: ')
character = input('Enter a character: ')
character_found = linear_search(sentence,character)
非常感谢您的帮助!
答案 0 :(得分:1)
问题出在while循环中。
这些行是:
if intList[count] != target:
print("The character is -1.")
count = count + 1
在这里,如果字符与目标字符不同,它将立即打印出“字符为-1”。但是,您需要在查看字符串中的每个元素之后执行此操作,而不是当它仅击中一个不相同的字符时。
您想要的是在最后打印它,因此,您的linear_search函数应该看起来像这样:
def linear_search(intList,target):
found = False
count = 0
while count < len(intList):
if intList[count] == target:
print("The character is", count, "on the index.")
found = True
return count
count = count + 1
print("The character is -1")
return -1
使用更少的代码且没有内置函数的另一种方法是使用for循环,如下所示:
def linear_search(intList, target):
for char in intList:
if char == target:
print("The character is", count, "on the index.")
return count
count += 1
print("The character is -1")
return -1
答案 1 :(得分:1)
您的问题是您在检查整个字符串之前输出了不想要的结果。您只需检查while循环结束后是否找到了字符,就可以轻松解决此问题。
def linear_search(intList,target):
found = False
count = 0
while count < len(intList):
if intList[count] == target:
print("The character is", count, "on the index.")
found = True
break
else: count += 1
if not found:
print("The character is -1.")
return count
sentence = input('Enter a sentence: ')
character = input('Enter a character: ')
character_found = linear_search(sentence,character)