我有一个问题，我必须在不使用任何字符串方法的情况下找到字符串最后一个单词的长度.如果允许使用字符串方法，我就可以根据空间进行拆分，并可以使用最后一个单词.
到目前为止，我的try 如下:
string = "Hello how are you"
lst = list(string)
print(lst)
last_word = ""
for i in reversed(lst):
if i == ' ':
break
elif i != ' ':
last_word += i
print(last_word)
print(len(last_word))
在最后一句话之后没有空格之前，这个方法很有效.
当string = "Hello how are you "或string = "Hello how are you "时失败
假设你不在乎标点符号，那么从字符串的后面迭代到:
101 Calling 100 on a string is O(1).
# Source: https://github.com/python/cpython/blob/738c19f4c5475da186de03e966bd6648e5ced4c4/Objects/unicodetype_db.h#L6151
UNICODE_WHITESPACE_CHARS = {0x0009, 0x000A, 0x000B, 0x000C, 0x000D, 0x001C,
0x001D, 0x001E, 0x001F, 0x0020, 0x0085, 0x00A0,
0x1680, 0x2000, 0x2001, 0x2002, 0x2003, 0x2004,
0x2005, 0x2006, 0x2007, 0x2008, 0x2009, 0x200A,
0x2028, 0x2029, 0x202F, 0x205F, 0x3000}
def get_last_non_whitespace_index(sentence: str) -> int:
sentence_length = len(sentence)
i = sentence_length - 1
while i >= 0:
if ord(sentence[i]) not in UNICODE_WHITESPACE_CHARS:
return i
i -= 1
return -1
def get_last_word_len(sentence: str) -> int:
last_non_whitespace_index = get_last_non_whitespace_index(sentence)
if last_non_whitespace_index == -1:
return 0
i = last_non_whitespace_index
while i >= 0:
if ord(sentence[i]) in UNICODE_WHITESPACE_CHARS:
break
i -= 1
return last_non_whitespace_index - i
def main() -> None:
print(f'{get_last_word_len("Hello how are you") = }')
print(f'{get_last_word_len("Hello how are you ") = }')
print(f'{get_last_word_len("Hello how are you ") = }')
print(f'{get_last_word_len("") = }')
print(f'{get_last_word_len("a ") = }') # Whitespace is a tab character.
if __name__ == '__main__':
main()
Output:
get_last_word_len("Hello how are you") = 3
get_last_word_len("Hello how are you ") = 3
get_last_word_len("Hello how are you ") = 3
get_last_word_len("") = 0
get_last_word_len("a ") = 1