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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