题目如下：

A character is unique in string S if it occurs exactly once in it.

For example, in string S = "LETTER", the only unique characters are "L" and "R".

Let's define UNIQ(S) as the number of unique characters in string S.

For example, UNIQ("LETTER") =  2.

Given a string S with only uppercases, calculate the sum of UNIQ(substring) over all non-empty substrings of S.

If there are two or more equal substrings at different positions in S, we consider them different.

Since the answer can be very large, return the answer modulo 10 ^ 9 + 7.

 

Example 1:

Input: "ABC"Output: 10Explanation: All possible substrings are: "A","B","C","AB","BC" and "ABC".Evey substring is composed with only unique letters.Sum of lengths of all substring is 1 + 1 + 1 + 2 + 2 + 3 = 10

Example 2:

Input: "ABA"Output: 8Explanation: The same as example 1, except uni("ABA") = 1.

 

Note: 0 <= S.length <= 10000.

解题思路：在任何子串中，只有出现一次的字符才对最终的结果起作用。假设输入的S为 XXXAXXXXAXXAXXXXX，X表示其他任意字符，现在我们来计算蓝A对最后的输出贡献了多少，很显然在两个红A之间的子串中，只要是包括蓝A的子串都有蓝A的贡献，如果第一个红A到蓝A之间的字符数量是L，蓝A到第二个红A之间的字符数量是R，那么蓝A的贡献就是 L + R + L*R + 1 ，其中L表示蓝A与左边字符组成的子串数量，R为与右边的，L*R为同时与左右结合，1表示不与任何字符结合。所有只有找出所有字符左右两边相同字符出现的位置，即可计算出最终的答案。

代码如下：

class Solution(object):    def uniqueLetterString(self, S):        """        :type S: str        :rtype: int        """        dic = {}        for i,v in enumerate(S):            dic[v] = dic.setdefault(v,[]) + [i]        res = 0        import bisect        for i,v in enumerate(S):            inx = bisect.bisect_left(dic[v], i)            before = i            after = len(S) - i - 1            if inx - 1 >= 0:                before = i - dic[v][inx - 1] - 1            if inx + 1 < len(dic[v]):                after = dic[v][inx + 1] - i - 1            res += (before + after + before * after + 1)        return res % (pow(10, 9) + 7)