打卡记录

统计子串中的唯一字符（找规律）

链接

大佬的题解

class Solution:
    def uniqueLetterString(self, s: str) -> int:
        ans = total = 0
        last0, last1 = {}, {}
        for i, c in enumerate(s):
            total += i - 2 * last0.get(c, -1) + last1.get(c, -1)
            ans += total
            last1[c] = last0.get(c, -1)
            last0[c] = i
        return ans