You are given a string?s?consisting of digits and an integer?k.

A?round?can be completed if the length of?s?is greater than?k. In one round, do the following:

1. Divide?s?into?consecutive groups?of size?k?such that the first?k?characters are in the first group, the next?k?characters are in the second group, and so on.?Note?that the size of the last group can be smaller than?k.

2. Replace?each group of?s?with a string representing the sum of all its digits. For example,?"346"?is replaced with?"13"?because?3 + 4 + 6 = 13.

3. Merge?consecutive groups together to form a new string. If the length of the string is greater than?k, repeat from step?1.

Return?s?after all rounds have been completed.

?

Example 1:

Input: s = "11111222223", k = 3

Output: "135"

Explanation:?

- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23". ?

Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. ?

?So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.?

- For the second round, we divide s into "346" and "5". ??

Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. ??

So, s becomes "13" + "5" = "135" after second round. Now, s.length <= k, so we return "135" as the answer.

Example 2:

Input: s = "00000000", k = 3

Output: "000"

Explanation:?

We divide s into "000", "000", and "00".?

Then we calculate the digit sum of each group:?

0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".

?需要2個循環(huán)，一個是每次字符串的長度>k，另一個是每次去用一個stringbuilder去存儲處理的到數(shù)據(jù)；

下面是代碼：

Constraints:

• 1 <= s.length <= 100

• 2 <= k <= 100

• s?consists of digits only.

Runtime:?1 ms, faster than?89.94%?of?Java?online submissions for?Calculate Digit Sum of a String.

Memory Usage:?40.7 MB, less than?73.27%?of?Java?online submissions for?Calculate Digit Sum of a String.