Table of Content
题目如下：
通过滑动窗口来取子字符串，并通过字典对象比较单词的出现次数可以求解这个问题。参考代码如下：
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class Solution: def findSubstring(self, s, words): """ :type s: str :type words: List[str] :rtype: List[int] """ # 初始化 strlen = 0 res = [] if strlen > len(s) or len(s) == 0 or len(words) == 0: return res dic_words = {} # 转换为字典,并计算出现次数 for w in words: if w in dic_words: dic_words[w] += 1 else: dic_words[w] = 1 # 子串长度 len_word = len(words[0]) len_words = len(words) strlen += len_word * len_words j = 0 # 子串滑动窗口 while j + strlen <= len(s): dic_substr = {} substr = s[j: j + strlen] k = 0 # 单词滑动窗口 while k + len_word <= len(substr): if substr[k: k + len_word] in dic_words: if substr[k: k + len_word] in dic_substr: dic_substr[substr[k: k + len_word]] += 1 else: dic_substr[substr[k: k + len_word]] = 1 k = k + len_word else: break if dic_words == dic_substr: res.append(j) j = j + 1 return res |
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