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$O(m\times n)$
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$O(m+n)$，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

class Solution(object):def strStr(self, haystack, needle):""":type haystack: str:type needle: str:rtype: int"""if len(needle) == 0:return 0if len(needle) > len(haystack):return -1next = self.getNext(needle)j = 0for i in range(len(haystack)):while j > 0 and haystack[i] != needle[j]:j = next[j-1]if haystack[i] == needle[j]:j += 1if j == len(needle):return i - j + 1return -1def getNext(self, needle):next = [0] * len(needle)j = 0for i in range(1, len(needle)):while j > 0 and needle[j] != needle[i]:j = next[j-1]if needle[j] == needle[i]:j += 1next[i] = jreturn nextclass 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 ireturn -1