剑指 Offer 48. 最长不含重复字符的子字符串

原题链接

begin为最长不含重复字符的子字符串的起点

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        begin,ans,dic = 0,0,{}
        for index,c in enumerate(s):
            if c in dic:
                begin = max(dic[c] + 1,begin) 
            dic[c] = index #更新字符c的坐标为当前坐标
            ans = max(index - begin + 1, ans)
        return ans