力扣leetcode30. 串联所有单词的子串

和《详细通俗的思路分析，多解法》解法相同，该文章已经写得很清晰了，就不再赘述，但是这个算法处理了很多不必要的位置。因为在最终匹配的解里面一定含有words[0]，那么先用O(n)时间（可以用kmp）找到所有的可能位置，然后对该位置使用滑动窗口即可，耗时为链接中算法的1/4左右，尽管在leetcode测试中笔者耗时为他的三倍多，但是数据量大才能有效地说明算法的效率问题

分别在1kw长度，20个words，words[0]长度1000，字符种类50
和1e长度，20个words，words[0]长度10000，字符种类50
下测的的结果，第一个为笔者的运行时间，单位s

代码写的比较乱，能知道增加部分的含义就行

 1 class Solution(object):
 2     def findSubstring(self, s, words):
 3         if not words:
 4             return []
 5         if not words[0]:
 6             return [i for i in range(len(s) + 1)]
 7         if len(s) < len(words) * len(words[0]):
 8             return []
 9 
10         set_word = {}
11         for i in words:
12             if i in set_word:
13                 set_word[i] += 1
14             else:
15                 set_word.update({i: 1})
16         len_word = len(words[0])
17         len_s = len(s)
18         words_num = len(words)
19 
20         arry = []
21         index = s.find(words[0])
22         while index != -1:
23             arry.append(index)
24             index = s.find(words[0], index + 1)
25         # print(arry)
26 
27         ans = set()
28         next_start = -1
29         for num in arry:
30             start = num - len_word * (words_num - 1)
31             if num >= next_start - 1:
32                 ans_temp, next_start = self.is_match(s, set_word, len_word, len_s, words_num, start, num)
33                 ans |= ans_temp
34 
35         # print(list(ans))
36         return list(ans)
37 
38     def is_match(self, s, set_word, len_word, len_s, words_num, start, end):
39         ans = set()
40         offset = 0
41         cache = set_word.copy()
42         word_list = []
43         next_start = -1
44         while start + (words_num - len(word_list)) * len_word <= len_s and start <= end + len(word_list) * len_word:
45             if start >= 0:
46                 same = 0
47                 for i in range(words_num - len(word_list)):
48                     this_word = ""
49                     for j in range(len_word):
50                         this_word += s[start + i * len_word + j]
51                     if this_word in cache:
52                         if cache[this_word] > 0:
53                             word_list.append(this_word)
54                             cache[this_word] -= 1
55                         elif this_word == word_list[0]:
56                             same = 1
57                             start += len_word * len(word_list)
58                             offset = len_word * (len(word_list) - 1)
59                             word_list.pop(0)
60                             cache[this_word] += 1
61                             break
62                         else:
63                             break
64                     else:
65                         break
66 
67                 if not same:
68                     flag = 0
69                     for i in cache:
70                         if cache[i] > 0:
71                             flag = 1
72                             break
73                     if flag:
74                         if word_list:
75                             for i in range(len(word_list)):
76                                 start += len_word
77                                 this_word = word_list.pop(0)
78                                 cache[this_word] += 1
79                         else:
80                             start += len_word
81                             cache = set_word.copy()
82                             offset = 0
83                     else:
84                         ans.add(start - offset)
85                         start += len_word * words_num
86                         offset = len_word * (words_num - 1)
87                         this_word = word_list.pop(0)
88                         cache[this_word] += 1
89                         next_start = start - offset
90             else:
91                 start += len_word
92         return ans, next_start

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原文地址：https://www.cnblogs.com/Pyrokine/p/12552519.html