30串联所有单词的子串

from typing import List
class Solution:
    def findSubstring(self, s: str, words: List[str]) -> List[int]:
        # 导入计数类
        from collections import Counter
        # 如果s和words其中一个为空，就返回空列表
        if not s or not words:return []
        res = []
        one_word = len(words[0])
        all_word = len(words) * one_word
        words = Counter(words)
        length = len(s)
        for i in range(0,length - all_word + 1):
            tmp = s[i:i + all_word]
            now_tmp = []
            for j in range(0,all_word,one_word):
                now_tmp.append(tmp[j:j + one_word])
            if Counter(now_tmp) == words:
                res.append(i)
        return res

A = Solution()
print(A.findSubstring("barfoothefoobarman",["foo","bar"]))
print(A.findSubstring("ababaab",["ab","ba","ba"]))

30串联所有单词的子串

原文：https://www.cnblogs.com/cong12586/p/13295137.html