Same code essentially, but more readable (in Python)


  • 0
    B

    I'd say the key observation is that all words are of the same length. Then it's guaranteed that by adding some offset, you can solve the problem in k passes of the text, given the length of each word is k.

    class Solution:
        # @param S, a string
        # @param L, a list of string
        # @return a list of integer
        def is_match(self, text, tokens, start, word_length, num_tokens):
            if len(text) - start < word_length * num_tokens:
                return False
            partial = collections.defaultdict(int)
            for j in xrange(start, start + num_tokens * word_length, word_length):
                    word = text[j:j+word_length]
                    if word in tokens:
                        partial[word] += 1
                        if partial[word] > tokens[word]:
                            return False
                    else:
                        return False
            return True
        def findSubstring(self, S, L):
            if not L:
                return []
            num_tokens, word_length = len(L), len(L[0])
            tokens = collections.Counter(L)
            results = []
            for offset in xrange(word_length):
                for i in xrange(offset, len(S), word_length):
                    if self.is_match(S, tokens, i, word_length, num_tokens):
                        results.append(i)
            return results
    

    By the way, the collections module of Python standard library is automatically imported for you, and it just saves you some codes with defaultdict and Counter.


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