[Leetcode]30. Substring with Concatenation of All Words @python

题目

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: “barfoothefoobarman”
words: [“foo”, “bar”]

You should return the indices: [0,9].
(order does not matter).

题目要求

给定一个字符串s和一个字符串列表words。找出字符串s中所有用words中的所有单词拼接的到的子串。参考给出的例子。

解题思路

这道题的解题思路参考南郭子綦.
用字典来统计words中每个单词出现的次数。因为words中的单词都是相同大小的，所以可以直接从s中截取固定大小的子串判读是否在words中出现。本实现应用了一个小的trick，用一个临时的字典来纪录每次匹配出现的word的数量，通过与words中每个单词出现的数量做对比提升代码的速度。

代码

class Solution(object):
    def findSubstring(self, s, words):
        """ :type s: str :type words: List[str] :rtype: List[int] """
        if len(words) == 0:
            return []
        wordsMap = {}
        for word in words:
            if word not in wordsMap:
                wordsMap[word] = 1
            else:
                wordsMap[word] += 1
        word_len = len(words[0])
        word_size = len(words)
        ans = []
        for i in range(len(s) - word_len * word_size + 1):
            j = 0
            cur_dict = {}
            while j < word_size:
                word = s[i + word_len * j:i + word_len * j + word_len]
                if word not in wordsMap:
                    break
                if word not in cur_dict:
                    cur_dict[word] = 1
                else:
                    cur_dict[word] += 1
                if cur_dict[word] > wordsMap[word]:
                    break
                j += 1
            if j == word_size:
                ans.append(i)
        return ans