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 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