原題地址:http://oj.leetcode.com/problems/implement-strstr/
題意:實現字符串匹配函數,並返回一個指針,這個指針指向原字符串中第一次出現待匹配字符串的位置。如:haystack='aabbaa'; needle='bb'。如果使用python實現,則最后返回的應該是一個字符串,即:'bbaa'。
解題思路:這道題我是使用KMP算法寫的,到現在KMP算法都不是很懂,只是按照《算法導論》上的偽代碼機械的實現了一遍。
代碼:
class Solution: # @param haystack, a string # @param needle, a string # @return a string or None # @KMP algorithms def ComputePrefixFunction(self, needle): Pi = [0 for i in range(len(needle))] m = len(needle) Pi[0] = 0 k = 0 for q in range(1, m): while k > 0 and needle[k] != needle[q]: k = Pi[k - 1] if needle[k] == needle[q]: k = k + 1 Pi[q] = k return Pi def strStr(self, haystack, needle): n = len(haystack) m = len(needle) if m == 0: return haystack Pi = self.ComputePrefixFunction(needle) q = 0 for i in range(0, n): while q > 0 and needle[q] != haystack[i]: q = Pi[q - 1] if needle[q] == haystack[i]: q = q + 1 if q == m: return haystack[i - m + 1:] return None
暴力匹配的算法,可以ac:
class Solution: # @param haystack, a string # @param needle, a string # @return a string or None def strStr(self, haystack, needle): if len(haystack) < len(needle): return None i = 0 while i < len(haystack)-len(needle)+1: j = 0; k = i while j < len(needle): if haystack[k] == needle[j]: j+=1; k+=1 else: break if j == len(needle): break else: i+=1 if i == len(haystack)-len(needle)+1: return None else: return haystack[i:]