You are given two string arrays,?queries?and?dictionary. All words in each array comprise of lowercase English letters and have the same length.

In one?edit?you can take a word from?queries, and change any letter in it to any other letter. Find all words from?queries?that, after a?maximum?of two edits, equal some word from?dictionary.

Return?a list of all words from?queries,?that match with some word from?dictionary?after a maximum of?two edits. Return the words in the?same order?they appear in?queries.

?

Example 1:

Input: queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]

Output: ["word","note","wood"]

Explanation:

- Changing the 'r' in "word" to 'o' allows it to equal the dictionary word "wood".?

- Changing the 'n' to 'j' and the 't' to 'k' in "note" changes it to "joke".?

- It would take more than 2 edits for "ants" to equal a dictionary word.

- "wood" can remain unchanged (0 edits) and match the corresponding dictionary word. Thus, we return ["word","note","wood"].

Example 2:

Input: queries = ["yes"], dictionary = ["not"]

Output: []

Explanation:

Applying any two edits to "yes" cannot make it equal to "not". Thus, we return an empty array.

?

Constraints:

• 1 <= queries.length, dictionary.length <= 100

• n == queries[i].length == dictionary[j].length

• 1 <= n <= 100

• All?queries[i]?and?dictionary[j]?are composed of lowercase English letters.

依次去遍歷2個字符串的差異字符數(shù)，如果小于等于2個，那么就放到list中，

返回即可；

Runtime:?3 ms, faster than?85.71%?of?Java?online submissions for?Words Within Two Edits of Dictionary.

Memory Usage:?42.1 MB, less than?99.40%?of?Java?online submissions for?Words Within Two Edits of Dictionary.