字符串最小表示法（LeetCode 899. Orderly Queue ）

A string S of lowercase letters is given.  Then, we may make any number of moves.

In each move, we choose one of the first K letters (starting from the left), remove it, and place it at the end of the string.

Return the lexicographically smallest string we could have after 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".

Note:

1. 1 <= K <= S.length <= 1000
2. S consists of lowercase letters only.

题解：

1、当K=1时，就是求字符串的最小表示法。

从i=0,j=1位置开始比较，当某个位置s[i+k]s[j+k]时，类似。

2、当K>=2，每次至少可以调整开头两个字符的位置。bazzz->  bzzza -> zzzab  ...->abzzz;

回想排序算法，只允许交换相邻的两个数字的位置，我们可以使序列有序。（每次把最大非有序的数字一直交换到最后相应位置）  

K=2时，54321->  32145  ->34521 ->21345  -> 23451 ->  12345  ;  每次我们调整最大的逆序的两个数字，依次可以把序列有序化。因此，K>=2时，最小系列就是排序得到的结果。

class Solution {
public:
    string get_min(string&s){
        int n=s.length();
        int i=0,j=1,k=0;
        while(i