思路： 没有。。。。真不会。。。。下次再来吧
代码：
def strStr(self, haystack: str, needle: str) -> int:
if not needle:
return 0
next = [0] * len(needle)
self.getNext(next, needle)
j = -1
for i in range(len(haystack)):
while j >= 0 and haystack[i] != needle[j+1]:
j = next[j]
if haystack[i] == needle[j+1]:
j += 1
if j == len(needle) - 1:
return i - len(needle) + 1
return -1
def getNext(self, next, s):
j = -1
next[0] = j
for i in range(1, len(s)):
while j >= 0 and s[i] != s[j+1]:
j = next[j]
if s[i] == s[j+1]:
j += 1
next[i] = j 459. 重复的子字符串
都要用到kmp算法了 还居然是简单难度！！！！！
不会
。。。。。。
代码：
def repeatedSubstringPattern(self, s: str) -> bool:
if len(s) == 0:
return False
nxt = [0] * len(s)
self.getNext(nxt, s)
if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0:
return True
return False
def getNext(self, nxt, s):
nxt[0] = -1
j = -1
for i in range(1, len(s)):
while j >= 0 and s[i] != s[j+1]:
j = nxt[j]
if s[i] == s[j+1]:
j += 1
nxt[i] = j
return nxt
