#### QUESTION:

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:

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. 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. 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".
```

**Constraints:**

```
1 <= s.length <= 100
2 <= k <= 100
s consists of digits only.
```

#### EXPLANATION:

比较简单, 直接一个for循环, 如果当前index是能被k整除, 那么说明需要添加到tmps上, 否则只需要加上就行. 比较关键的点是, 如果最后不是能整除的, 需要把last的也加上.

#### SOLUTION:

```
class Solution {
func digitSum(_ s: String, _ k: Int) -> String {
var str:String = s
while str.count > k {
var tmp:Int = 0
var tmpS:String = ""
for i in 0...str.count - 1 {
tmp += Int(String(Array(str)[i]))!
if (i + 1) % k == 0 {
tmpS += String(tmp)
tmp = 0
}
}
if str.count % k != 0 {
tmpS += String(tmp)
}
str = tmpS
}
return str
}
}
```