30. 串联所有单词的子串
难度:困难
给定一个字符串 s
和一些 长度相同 的单词 words
**。**找出 s
中恰好可以由 words
中所有单词串联形成的子串的起始位置。
注意子串要与 words
中的单词完全匹配,中间不能有其他字符 ,但不需要考虑 words
中单词串联的顺序。
示例 1:
输入:s = "barfoothefoobarman", words = ["foo","bar"]
输出:[0,9]
解释:
从索引 0 和 9 开始的子串分别是 "barfoo" 和 "foobar" 。
输出的顺序不重要, [9,0] 也是有效答案。
示例 2:
输入:s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
输出:[]
示例 3:
输入:s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
输出:[6,9,12]
提示:
1 <= s.length <= 104
-
s
由小写英文字母组成 1 <= words.length <= 5000
1 <= words[i].length <= 30
-
words[i]
由小写英文字母组成
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
word_len = len(words[0])
word_num = len(words)
words_len = word_len * word_num
n = len(s)
word_counter = Counter(words)
i = 0
result = []
while i < n - words_len + 1:
j = i
temp_counter = []
for _ in range(word_num):
temp_counter.append(s[j:j+word_len])
j += word_len
if Counter(temp_counter) == word_counter:
result.append(i)
i += 1
return result