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 "barfoor" 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：[]
暴力 2000ms
class Solution:
def findSubstring(self, s, words):
"""
:type s: str
:type words: List[str]
:rtype: List[int]
"""
if len(words) == 0 or len(s) == 0:
return []
n = len(words)
len_word = len(words[0])
len_words = len("".join(words))
len_s = len(s)
res = []
for i in range(len_s - len_words + 1):
window = s[i:i+len_words]
l = []
for word in words:
l.append(word)
for j in range(n):
word = window[j*len_word:(j+1)*len_word]
if word in l:
l.remove(word)
else:
break
if l == []:
res.append(i)
return res
64ms
class Solution:
def findSubstring(self, s, words):
if len(words) == 0:
return []
lens = len(s)
lenw = len(words[0])
lenws = lenw * len(words)
if lens < lenws:
return []
counter = {}
for i in range(len(words)):
if words[i] in counter:
counter[words[i]] += 1
else:
counter[words[i]] = 1
res = []
for i in range(min(lenw, lens-lenws + 1)):
s_pos = word_pos = i
d = {}
while s_pos + lenws <= lens:
# 截取单词
word = s[word_pos:word_pos + lenw]
# 移动到下一个单词
word_pos += lenw
if word not in counter:
s_pos = word_pos
d.clear()
else:
if word not in d:
d[word] = 1
else:
d[word] += 1
while d[word] > counter[word]:
d[s[s_pos:s_pos + lenw]] -= 1
s_pos += lenw
if word_pos - s_pos == lenws:
res.append(s_pos)
return res
