Leetcode #28. Implement strStr()

Implement strStr().

Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

 

Brute Force算法，时间复杂度 O(mn)

 

 1 class Solution(object):
 2     def strStr(self, haystack, needle):
 3         """
 4         :type haystack: str
 5         :type needle: str
 6         :rtype: int
 7         """
 8         m = len(haystack)
 9         n = len(needle)
10         
11         if n == 0:
12             return 0
13         
14         if m < n:
15             return -1
16             
17         
18         for i in range(m-n+1):
19             if haystack[i:i+n] == needle:
20                 return i
21 
22         return -1

 

 

Rabin karp算法时间复杂度可以降低到 O(mn) on average.

haystack: abcdefgh, needle: abc

needle_code = a + b*p + c*p^2 使用sliding window计算haystack_code可以节省计算量。

KMP算法也很快，但是面试一般无法完成。

 1 def charToInt(c):                                                                                                   
 2     return ord(c) - ord('a') + 1                                                                                    
 3                                                                                                                     
 4 def strStrRK(haystack, needle):                                                                                     
 5                                                                                                                     
 6     m = len(haystack)                                                                                               
 7     n = len(needle)                                                                                                 
 8                                                                                                                     
 9     if n == 0:                                                                                                      
10         return 0                                                                                                    
11     if m < n:                                                                                                       
12         return -1                                                                                                   
13                                                                                                                     
14     #choose a prime number as base                                                                                  
15     base = 29                                                                                                       
16                                                                                                                     
17     needle_code = 0                                                                                                 
18     for i in range(n):                                                                                              
19         needle_code += charToInt(needle[i]) * base**i                                                               
20                                                                                                                     
21                                                                                                                     
22     lead = charToInt(haystack[0])                                                                                   
23     for j in range(m - n + 1):                                                                                      
24                                                                                                                     
25         if j == 0:                                                                                                  
26             haystack_code = 0                                                                                       
27             for i in range(n):                                                                                      
28                 haystack_code += charToInt(haystack[j + i])*base**i                                                 
29         else:                                                                                                       
30             haystack_code -= lead                                                                                   
31             haystack_code /= base                                                                                   
32             haystack_code += charToInt(haystack[j + n - 1])*base**(n-1)                                             
33             lead = charToInt(haystack[j])                                                                           
34                                                                                                                     
35         if haystack_code == needle_code:                                                                            
36             if haystack[j:j + n] == needle:                                                                         
37                 return j                                                                                            
38                                                                                                                     
39     return -1

 

posted @ 2016-11-27 09:32  lettuan  阅读(132)  评论(0)  编辑  收藏  举报