这是字符串匹配问题,朴素做法是两重遍历,依次从主串的i位置开始查看是否和模式串匹配,若不匹配就换下一个位置进行判断,直到找到或者遍历完,时间复杂度 O ( m × n ) O(m \times n) O(m×n)
还可以对主串进行处理,把所有匹配模式串的字串替换为"1",然后在替换后的主串里面寻找第一个"1"出现的位置
但是这道题时间复杂度最低的还是得考虑用KMP算法,时间复杂度 O ( m + n ) O(m + n) O(m+n),这里我是看到Youtube上一位博主的讲解才恍然大悟的,链接在这里
python
class Solution(object):
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
if len(needle) == 0:
return 0
if len(needle) > len(haystack):
return -1
next = self.getNext(needle)
j = 0
for i in range(len(haystack)):
while j > 0 and haystack[i] != needle[j]:
j = next[j-1]
if haystack[i] == needle[j]:
j += 1
if j == len(needle):
return i - j + 1
return -1
def getNext(self, needle):
next = [0] * len(needle)
j = 0
for i in range(1, len(needle)):
while j > 0 and needle[j] != needle[i]:
j = next[j-1]
if needle[j] == needle[i]:
j += 1
next[i] = j
return next
class Solution(object):
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
restr = haystack.replace(needle, '1')
for i in range(len(restr)):
if restr[i] == '1':
return i
return -1