You are given an array of equal-length strings?words. Assume that the length of each string is?n.

Each string?words[i]?can be converted into a?difference integer array?difference[i]?of length?n - 1?where?difference[i][j] = words[i][j+1] - words[i][j]?where?0 <= j <= n - 2. Note that the difference between two letters is the difference between their?positions?in the alphabet i.e.?the position of?'a'?is?0,?'b'?is?1, and?'z'?is?25.

• For example, for the string?"acb", the difference integer array is?[2 - 0, 1 - 2] = [2, -1].

All the strings in words have the same difference integer array,?except one. You should find that string.

Return?the string in?words?that has different?difference integer array.

?

Example 1:

Input: words = ["adc","wzy","abc"]Output: "abc"Explanation: - The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1]. - The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1]. - The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1]. The odd array out is [1, 1], so we return the corresponding string, "abc".

Example 2:

Input: words = ["aaa","bob","ccc","ddd"]Output: "bob"Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].

?

Constraints:

• 3 <= words.length <= 100

• n == words[i].length

• 2 <= n <= 20

• words[i]?consists of lowercase English letters.

hashmap可以做list跟string的映射，之前都沒(méi)有試過(guò)，第一次嘗試，用起來(lái)還可以，如果這樣的話，

可以判斷重復(fù)的list，

Runtime3 ms

Beats

60.75%

Memory39.9 MB

Beats

93.18%