Leetcode 899. An orderly queue

Given a string s and an integer k .You can choose one of the first k letters of s and add it to the end of the string.

returns the lexicographically smallest string after applying any number of moves from the above steps.

Example 1:

Input: s = "cba", k = 1
Output: "acb"
Explanation:
In the first step, we move the first character ("c") to the end, to get the string "bac".
In the second step, we move the first character ("b") to the end to get the final result "acb".
Example 2:

Input: s = "baaca", k = 3
Output: "aaabc"
Explanation:
In the first step, we move the first character ("b") to the end, to get the string "aacab".
In the second step, we move the third character ("c") to the end to get the final result "aaabc".

Tip:

1 <= k <= S.length <= 1000
s consists of lowercase letters only.

Thinking:

When k==1, the original string is regarded as a ring, the first character is placed at the end every time from the beginning to the end, and the lexicographical order among these characters is saved and simulated directly.

When k>=2, after a limited number of operations, it must be in ascending order, because we can choose the largest one in the first k every time and put it at the end, which is very troublesome in practice, but withThe result of sorting is the same, so you can return directly to sort.

class Solution {public:string orderlyQueue(string s, int k) {if(k>1){sort(s.begin(),s.end());return s;}string ans=s;for(int i=0;i<s.size();++i){s=string{s.begin()+1,s.end()}+s[0];span>ans=min(ans,s);}return ans;}};