题目描述:
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:[]
解题思路:
class Solution:
def findSubstring(self, s: str, words: List[str]) -> List[int]:
cnt = len(words[0])
length = len(words) * cnt
head = 0
tail = head + length
output = []
has = True
while(tail <= len(s)):
words_tmp = words[:]
tmp = s[head:tail]
while(tmp):
if tmp[:cnt] in words_tmp:
words_tmp.remove(tmp[:cnt])
tmp = tmp[cnt:]
has = True
else:
has = False
break
if has == True:
output.append(head)
head = head + 1
tail = head + length
return output
算法性能: