即abc输出abc,acb,bac,bca,cab,cba

思路:先以非递归方式,完成部分

public static void Rotate(char[] str){    var length = str.Length;    int i = 1;    for (int j = 0; j < length; j++)    {                foreach (var item in str)        {            Console.Write(item);        }        Console.WriteLine();        char temp = str[0];        //move left        for (i = 1; i < length; i++)        {            str[i - 1] = str[i];        }        str[i - 1] = temp;    }}

得到结果是abc,bca,cab

每个数右边的n-1个数可以进行排列abc,acb.实际上就是对bc进行Rotate.

递归退出点就是输出点

public static void Rotate(char[] str,int pos){    var length = str.Length;    if (pos == (length - 1))    {        foreach (var item in str)        {            Console.Write(item);        }        Console.WriteLine();        return;    }     int i = 1;    for (int j = pos; j < length; j++)    {        Rotate(str, pos+1);        char temp = str[pos];        for (i = pos + 1; i < length; i++)        {            str[i - 1] = str[i];        }        str[i - 1] = temp;    }}

未完