力扣leetcode30 串聯所有單詞的子串

和《詳細通俗的思路分析，多解法》解法相同，該文章已經寫得很清晰了，就不再贅述，但是這個演算法處理了很多不必要的位置。因為在最終匹配的解裡面一定含有words[0]，那麼先用o(n)時間（可以用kmp）找到所有的可能位置，然後對該位置使用滑動視窗即可，耗時為鏈結中演算法的1/4左右，儘管在leetcode測試中筆者耗時為他的三倍多，但是資料量大才能有效地說明演算法的效率問題

分別在1kw長度，20個words，words[0]長度1000，字元種類50

和1e長度，20個words，words[0]長度10000，字元種類50

下測的的結果，第乙個為筆者的執行時間，單位s

**寫的比較亂，能知道增加部分的含義就行

1
class
solution(object):
2def
findsubstring(self, s, words):3if
notwords:
4return
5ifnotwords[0]:
6return [i for i in range(len(s) + 1)]
7if len(s) < len(words) *len(words[0]):
8return
910         set_word ={}
11for i in
words:
12if i in
set_word:
13                 set_word[i] += 1
14else
:15                 set_word.update()
16         len_word =len(words[0])
17         len_s =len(s)
18         words_num =len(words)
1920         arry =
21         index =s.find(words[0])
22while index != -1:
2324             index = s.find(words[0], index + 1)25#
print(arry)
2627         ans =set()
28         next_start = -1
29for num in
arry:
30             start = num - len_word * (words_num - 1)
31if 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
3435
#print(list(ans))
36return
list(ans)
3738
defis_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
44while start + (words_num - len(word_list)) * len_word <= len_s and start <= end + len(word_list) *len_word:
45if start >=0:
46                 same =0
47for i in range(words_num -len(word_list)):
48                     this_word = ""
49for j in
range(len_word):
50                         this_word += s[start + i * len_word +j]
51if this_word in
cache:
52if cache[this_word] >0:
5354                             cache[this_word] -= 1
55elif this_word ==word_list[0]:
56                             same = 1
57                             start += len_word *len(word_list)
58                             offset = len_word * (len(word_list) - 1)
59word_list.pop(0)
60                             cache[this_word] += 1
61break
62else:63
break
64else:65
break
6667
ifnot
same:
68                     flag =0
69for i in
cache:
70if cache[i] >0:
71                             flag = 1
72break
73if
flag:
74if
word_list:
75for i in
range(len(word_list)):
76                                 start +=len_word
77                                 this_word =word_list.pop(0)
78                                 cache[this_word] += 1
79else
:80                             start +=len_word
81                             cache =set_word.copy()
82                             offset =0
83else
: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
90else
:91                 start +=len_word
92return ans, next_start

力扣leetcode30 串聯所有單詞的子串

力扣 LeetCode 30 串聯所有單詞的子串

leetcode 30 串聯所有單詞的子串

leetcode 30 串聯所有單詞的子串

力扣leetcode30 串聯所有單詞的子串

力扣 LeetCode 30 串聯所有單詞的子串

leetcode 30 串聯所有單詞的子串

leetcode 30 串聯所有單詞的子串

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