1+ """
2+ Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
3+
4+ Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
5+
6+ The order of output does not matter.
7+
8+ Example 1:
9+
10+ Input:
11+ s: "cbaebabacd" p: "abc"
12+
13+ Output:
14+ [0, 6]
15+
16+ Explanation:
17+ The substring with start index = 0 is "cba", which is an anagram of "abc".
18+ The substring with start index = 6 is "bac", which is an anagram of "abc".
19+ Example 2:
20+
21+ Input:
22+ s: "abab" p: "ab"
23+
24+ Output:
25+ [0, 1, 2]
26+
27+ Explanation:
28+ The substring with start index = 0 is "ab", which is an anagram of "ab".
29+ The substring with start index = 1 is "ba", which is an anagram of "ab".
30+ The substring with start index = 2 is "ab", which is an anagram of "ab".
31+
32+
33+ 给定一个字符串 s 和 非空字符串 p。 找到所有 p 在 s 中的变位词,将变位词开始部分的索引添加到结果中。
34+
35+ 字符串包含的字符均为小写英文字母。
36+
37+ 结果的顺序无关紧要。
38+
39+
40+ 思路:
41+ 看到标的是个 easy .. 估计数据不会很大,直接 sorted(s[i:i+len(p)]) == sorted(p) 判断了..
42+ 结果是 TLE。
43+
44+ 好吧,只能认真思考下了。
45+
46+ 想到的是利用动态规划,还有点回溯的味道。
47+
48+ 子问题为:
49+ 当前点加上之前的点是否覆盖了全部的p。
50+
51+ 若全部覆盖了,则返回 i - len(p) + 1 的开始索引,同时,将 s[i - len(p) + 1] 处的字符添加到 dp 的判断点中。
52+ 这样做可以扫描到这种情况:
53+ s = ababab
54+ p = ab
55+ 循环这种。
56+
57+ 若没有存在于最后一个dp中,那么先判断是否有重复:
58+ 1. 如果在 p 中,但不在上一个子问题的需求解里。
59+ 2. 如果与最前面的一个字符发生了重复。
60+ 3. 如果与之前的一个字符不同。
61+ 满足这三个条件那么可以无视这个重复的字符。
62+ 比如这样:
63+ s = abac
64+ p = abc
65+
66+ 这样相当于把重复的一个 a 忽略掉了。
67+ 同时还要判断是否它的前一个是否一致,一致的话就不能单纯的忽略了。
68+
69+ s = abaac / aabc
70+ p = abc
71+
72+ 没有满足的上述三个条件,那么重新复制一份原始子问题,然后将当前字符判断一下,继续下一个循环即可。
73+
74+ beat 99% 84ms
75+ 测试地址:
76+ https://leetcode.com/problems/find-all-anagrams-in-a-string/description/
77+
78+ """
79+ class Solution (object ):
80+ def findAnagrams (self , s , p ):
81+ """
82+ :type s: str
83+ :type p: str
84+ :rtype: List[int]
85+ """
86+ 87+ # p = sorted(p)
88+ if not s :
89+ return []
90+ _p = {}
91+ 92+ for i in p :
93+ try :
94+ _p [i ] += 1
95+ except :
96+ _p [i ] = 1
97+ 98+ result = []
99+ 100+ x = _p .copy ()
101+ if s [0 ] in _p :
102+ x [s [0 ]] -= 1
103+ if not x [s [0 ]]:
104+ x .pop (s [0 ])
105+ 106+ dp = x
107+ if not dp :
108+ return [i for i in range (len (s )) if s [i ] == p ]
109+ 110+ for i in range (1 , len (s )):
111+ if s [i ] in dp :
112+ t = dp
113+ t [s [i ]] -= 1
114+ if not t [s [i ]]:
115+ t .pop (s [i ])
116+ 117+ if not t :
118+ result .append (i - len (p )+ 1 )
119+ x = {}
120+ _t = s [i - len (p )+ 1 ]
121+ x [_t ] = 1
122+ 123+ dp = x
124+ else :
125+ if s [i ] in _p and s [i ] != s [i - 1 ] and s [i ] == s [i - len (p )+ 1 ]:
126+ continue
127+ x = _p .copy ()
128+ if s [i ] in x :
129+ x [s [i ]] -= 1
130+ if not x [s [i ]]:
131+ x .pop (s [i ])
132+ dp = x
133+ 134+ return result
135+
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