最简单的字符串匹配方法,传说中的在特殊情况的暴力求解:
伪代码:
naive_string_matcher(t,p):
n=len(t)
m=len(p)
for s =0 to n-m:
if p[1..m]==t[s+1..s+m]:
print("success!")
具体的代码:
str1 = 'aabacgaabgcaab'
str2 = 'aab'
def naive_string_match(str1, str2):
n1 = len(str1) # 8
n2 = len(str2) # 3
result = []
index = 0
for s in range(n1 - n2 + 1): # 0 1 2
# print(s)
for j in range(n2):
if str1[s + j] == str2[j]:
index += 1
else:
index = 0
break
if index == n2:
result.append(s)
return result
print("返回匹配点的下标:")
print(naive_string_match(str1, str2))
运行结果
返回匹配点的下标:
[0, 6, 11]