substring-with-concatenation-of-all-words

最新推荐文章于 2019-08-24 22:07:23 发布

原创最新推荐文章于 2019-08-24 22:07:23 发布 · 305 阅读

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#算法 python #string

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本文介绍了一种寻找特定模式字符串在目标字符串中所有起始位置的高效算法。该算法通过滑动窗口的方式，在目标字符串中查找由指定单词列表组合而成的子串，并返回这些子串的所有可能起始索引。

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

For example, given:
S:"barfoothefoobarman"
L:["foo", "bar"]

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

def findSubstring(S, L):
   map = {}
   for i in L:
      if i not in map:
         map[i] = 1
      else:
         map[i] += 1
   begin = 0
   end = 0
   le = len(L[0])
   count = len(L)
   res = []
   while end < len(S)-le + 1:
      if S[end:end+le] not in map:
         map[S[end:end + le]] = -1
      if map[S[end:end + le]] > 0:
         map[S[end:end + le]] -= 1
         count -= 1
         end = end + le
      else:
         end += 1
      while count == 0:
         if map[S[begin:begin+le]] == 0:
            res.append(begin)
            begin = end
            count = len(L)
            for i in L:
               if i not in map:
                  map[i] = 1
               else:
                  map[i] += 1
         else:
            begin += 1
   return res
a = ["foo", "bar","the"]
print findSubstring("barfoothefoobarmanthe", a)

class S:
    def __init__(self):
        pass

    @staticmethod
    def find_substring(s, words):
        if len(s) < len(words)*len(words[0]):
            return []

        word_len, word_num = len(words[0]), len(words)
        word_dic = {}
        for i in words:
            if i not in word_dic:
                word_dic[i] = 1
            else:
                word_dic[i] += 1

        result = []
        for i in range(len(s)+1-word_len*word_num):
            tmp_dic, j = {}, 0
            while j < word_num:
                word = s[i+j*word_len:i+j*word_len+word_len]
                if word not in word_dic:
                    break

                if word not in tmp_dic:
                    tmp_dic[word] = 1
                else:
                    tmp_dic[word] += 1

                if tmp_dic[word] > word_dic[word]:
                    break

                j += 1
                if j == word_num:
                    result.append(i)

        return result


s = S()
print(s.find_substring('ashjkdfghjkl', ['dfg', 'hjk']))