# 899. Orderly Queue

Easy
You are given a string `s` and an integer `k`. You can choose one of the first `k` letters of `s` and append it at the end of the string..
Return the lexicographically smallest string you could have after applying the mentioned step any number of moves.
Example 1:
Input: s = "cba", k = 1
Output:
"acb"
Explanation:
In the first move, we move the 1st character 'c' to the end, obtaining the string "bac".
In the second move, we move the 1st character 'b' to the end, obtaining the final result "acb".
Example 2:
Input: s = "baaca", k = 3
Output:
"aaabc"
Explanation:
In the first move, we move the 1st character 'b' to the end, obtaining the string "aacab".
In the second move, we move the 3rd character 'c' to the end, obtaining the final result "aaabc".
Constraints:
• `1 <= k <= s.length <= 1000`
• `s` consist of lowercase English letters.

### 解題

k 等於 1 的時候只能夠不斷把第一個字母挪到屁股，回傳其中最小的字串。而 k 大於等於2的時候，可以排列出所有組合，所以直接用 sort 公式排序回傳即可。
Runtime: 0 ms, faster than 100%
Memory Usage: 4.9 MB, less than 60%
func orderlyQueue(s string, k int) string {
res := s
if k == 1 {
for i := 0; i < len(s); i++ {
res = min(s[1:] + s[:1], res)
s = s[1:] + s[:1]
}
} else {
bytes := []byte(s)
sort.Slice(bytes, func(i int, j int) bool { return bytes[i] < bytes[j] })
res = string(bytes)
}
return res
}
func min(a, b string) string {
if a < b { return a }
return b
}