leedcode 找出字符串中第一个匹配项的下标

自己写的

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        haystack_len = len(haystack)
        needle_len = len(needle)

        if haystack == needle:
            return 0

        if haystack_len < needle_len:
            return -1
        i = 0

        while i <= haystack_len - needle_len:
            if haystack[i] == needle[0]:  # 遍历haystack如果和needle第一个相同就开始对比

                if needle_len==1:
                    return i
                j = 1
                while j < needle_len:
                    if haystack[i + j] == needle[j]:

                        j += 1

                        if j == needle_len:
                            return i
                    else:
                        i += 1
                        if i == haystack_len - needle_len + 1:
                            return -1
                        break

            else:  # 遍历haystack如果和needle第一个不相同就下一个
                i += 1
                if i == haystack_len - needle_len + 1:
                    return -1

gpt回答

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        haystack_len = len(haystack)
        needle_len = len(needle)

        if needle_len == 0:
            return 0

        for i in range(haystack_len - needle_len + 1):
            if haystack[i:i + needle_len] == needle:
                return i

        return -1

class Solution:
    def strStr(self, haystack: str, needle: str) -> int:
        haystack_len = len(haystack)
        needle_len = len(needle)

        if needle_len == 0:
            return 0

        for i in range(haystack_len - needle_len + 1):
            # 截取 haystack 中与 needle 长度相同的子串
            current_substring = haystack[i:i + needle_len]
            
            # 检查子串是否与 needle 相等
            if current_substring == needle:
                return i

        # 如果没有找到匹配，返回 -1
        return -1