You are given two strings order and s. All the words of?order?are?unique?and were sorted in some custom order previously.

Permute the characters of?s?so that they match the order that?order?was sorted. More specifically, if a character?x?occurs before a character?y?in?order, then?x?should occur before?y?in the permuted string.

Return?any permutation of?s?that satisfies this property.

?

Example 1:

Input: order = "cba", s = "abcd"Output: "cbad"Explanation: "a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a". Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.

Example 2:

Input: order = "cbafg", s = "abcd"Output: "cbad"

?

Constraints:

• 1 <= order.length <= 26

• 1 <= s.length <= 200

• order?and?s?consist of lowercase English letters.

• All the characters of?order?are?unique.

先將字符串變成字符數(shù)組，然后根據(jù)排序的字符串依次append，就可以了，挺好的。

Runtime:?0 ms, faster than?100.00%?of?Java?online submissions for?Custom Sort String.

Memory Usage:?40 MB, less than?89.76%?of?Java?online submissions for?Custom Sort String.