[LeetCode]30 串联所有单词的子串

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 "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：[]

代码

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class Solution:
    def findSubstring(self, s, words):
        """
        :type s: str
        :type words: List[str]
        :rtype: List[int]
        """
        if len(words) == 0:
            return []
        
        str_len = len(s)
        len_word = len(words[0])
        total_len = len_word * len(words)
        if str_len < total_len:
            return []

        words_dic = {}
        for word in words:
            words_dic[word] = words_dic.get(word, 0) + 1
            
        result = []
        for str_start in range(min(len_word, str_len - total_len + 1)):
            head = str_start
            tail = head + len_word
            scan_dic = {}
            while tail <= str_len:
                cur_word = s[tail - len_word:tail]
                if cur_word in words_dic:
                    scan_dic[cur_word] = scan_dic.get(cur_word, 0) + 1
                    while scan_dic[cur_word] > words_dic[cur_word]:
                        scan_dic[s[head:head + len_word]] -= 1
                        head += len_word
                    if tail - head == total_len:
                        result.append(head)
                        scan_dic[s[head:head + len_word]] -= 1
                        head += len_word
                else:
                    scan_dic.clear()
                    head = tail
                tail += len_word
                    
        return result