You are given a string?s, a string?chars?of?distinct?characters and an integer array?vals?of the same length as?chars.

The?cost of the substring?is the sum of the values of each character in the substring. The cost of an empty string is considered?0.

The?value of the character?is defined in the following way:

• If the character is not in the string?chars, then its value is its corresponding position?(1-indexed)?in the alphabet.

• For example, the value of?'a'?is?1, the value of?'b'?is?2, and so on. The value of?'z'?is?26.

• Otherwise, assuming?i?is the index where the character occurs in the string?chars, then its value is?vals[i].

Return?the maximum cost among all substrings of the string?s.

?

Example 1:

Input: s = "adaa", chars = "d", vals = [-1000]

Output: 2

Explanation: The value of the characters "a" and "d" is 1 and -1000 respectively. The substring with the maximum cost is "aa" and its cost is 1 + 1 = 2. It can be proven that 2 is the maximum cost.

Example 2:

Input: s = "abc", chars = "abc", vals = [-1,-1,-1]

Output: 0

Explanation: The value of the characters "a", "b" and "c" is -1, -1, and -1 respectively. The substring with the maximum cost is the empty substring "" and its cost is 0. It can be proven that 0 is the maximum cost.

?建一個(gè)hashmap，把chars的字符對應(yīng)的值放到map中，

然后遍歷字符串，如果在map中，將對應(yīng)的值放到數(shù)組中，如果沒在，就-‘a(chǎn)’+1 返回?cái)?shù)組，

最后遍歷數(shù)組判斷最大值，但是如果sum小于0了，就沒必要繼續(xù)保留了，直接sum=0，繼續(xù)遍歷剩下的數(shù)組。

返回最大值即可；

Constraints:

• 1 <= s.length <= 105

• s?consist of lowercase English letters.

• 1 <= chars.length <= 26

• chars?consist of?distinct?lowercase English letters.

• vals.length == chars.length

• -1000 <= vals[i] <= 1000

Runtime:?24 ms, faster than?100.00%?of?Java?online submissions for?Find the Substring With Maximum Cost.

Memory Usage:?43.5 MB, less than?100.00%?of?Java?online submissions for?Find the Substring With Maximum Cost.