递归变位数(练习)

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