递归--变位数(练习)
2011-09-15 23:01 Clingingboy 阅读(565) 评论(0) 编辑 收藏 举报
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