[Leetcode][Python]30: Substring with Concatenation of All Words

# -*- coding: utf8 -*-‘‘‘__author__ = ‘[email protected]‘

30: Substring with Concatenation of All Wordshttps://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/

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 Lexactly 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).

===Comments by Dabay===

生成一个hash表来记录每个词在L中出现的次数。扫描要检查的字符串S，检查以这个字母开始的 长度为L总长度 的字符串，如果小词在hash表中，值减一；如果最后每一个hash值都是0，说明正好全部匹配完，加入到结果。重新初始化hash表，进行下一次检查。‘‘‘

class Solution:    # @param S, a string    # @param L, a list of string    # @return a list of integer    def findSubstring(self, S, L):        d = {}        for x in L:            if x not in d:                d[x] = 1            else:                d[x] += 1        res = []        word_length = len(L[0])        i = 0        while i < len(S) - word_length * len(L) + 1:            j = i            dd = dict(d)            while j < i + word_length * len(L):                word = S[j:j+word_length]                if word in dd:                    dd[word] -= 1                    if dd[word] < 0:                        break                else:                    break                j += word_length            else:                res.append(i)            i += 1        return res

def main():    s = Solution()    string = "aaa"    dictionary = ["a", "b"]    print s.findSubstring(string, dictionary)

if __name__ == "__main__":    import time    start = time.clock()    main()    print "%s sec" % (time.clock() - start)