[LeetCode]题解（python）：030-Substring with Concatenation of All Words

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题目来源

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https://leetcode.com/problems/substring-with-concatenation-of-all-words/

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 wordsexactly once and without any intervening characters.

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题意分析

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Input：a str and a list

Output:a list recorded the indices

Conditions：输入一个字符串s和一连串的长度相同的字符串数组words，找出仅由所有的words组成的s的子字符串起始位置。words内的元素可能重复，但是每个元素都需要出现（如有两个‘good’元素，那么’good‘就需要出现两次）

examples:

s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].

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题目思路

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1. 首先将words存为一个dic（dictiionary（字典））（出现一次value为1，出现两次value为2，以此类推）
2. 遍历每一个可能的str（长度等于len（words）*len（words[0]））
3. 对每一个str，用一个新的字典d来存储已经出现的word，用python的分片来访问每一个str里面的word，用计数器count计数相等的word
4. 条件检验：如果元素不在dic，直接break取下一个可能的str；如果元素在dic，但是不在d，直接将新元素加入到d，value取为1，count加1；如果元素在dic，同时也在d，看此时d中的value是否小于dic中的value，小于则value加1，count加1，否则此字串不满足条件，舍弃取下一条str
5. 每一次对一个str处理后，d要清空，count要置为0
6. 注意事项：d.get(each_str, -1) != -1的速度不如 each_str in d

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AC代码（Python）

 _author_ = "YE"
 # -*- coding:utf-8 -*-
 # SH……把d.get(each_str, -1) == -1类似的判断改为 each_str not in d 就AC了……说明后者效率高呀
 def findSubstring(s, words):
         """
         :type s: str
         :type words: List[str]
         :rtype: List[int]
         """
         import math
         rtype = []
         dic = {}

         len_word = len(words)
         if len_word == 0:
             return rtype
         #print('len of words:%s' % len_word)

         len_each_word = len(words[0])
         #print('len of each word:%s' % len_each_word)

         all_len = len_word * len_each_word
         #print("all len:", all_len)

         for x in words:
             if x  in dic:
                 dic[x] = dic[x] + 1
             else:
                 dic[x] = 1

         #print('dictionary:')
         #print(dic)

         #print(dic.get('arf',-1) == -1)

         len_s = len(s)
         #print('the len of the str s', len_s)

         if len_s < all_len:
             return rtype

         #print('From loop')

         for i in range(len_s - all_len + 1):
             str = s[i:i+all_len]
             count = 0
             d = {}
             for j in range(len_word):
                 each_str = str[j * len_each_word:(j + 1) * len_each_word]
                 if each_str not in dic:
                     count = 0
                     break
                 elif each_str in d:
                     if d[each_str] == dic[each_str]:
                         count = 0
                         break
                     else:
                         d[each_str] = d[each_str] + 1
                         count = count + 1
                         continue
                 else:
                     d[each_str] = 1
                     count = count + 1
             if count == len_word:
                 rtype.append(i)
                 count = 0
         #print(rtype)