计算java中的递归步骤

Question

我想计算递归步数,并在达到某个限制时停止递归.

实际上我正在处理河内塔问题,我想限制为解决问题而执行的幻灯片数量.这是我的解决方案:

class HanoiNK{

    public static void main(String args[]){

            int n = 4;
            int k = 5;

            try{
                slide(k, n, 'A', 'B', 'C');
            }catch(Exception e){
                System.out.println(e);
            }
    }

    public static void slide(int counter, int height, char source,
                              char buffer, char destination) throws Exception{      
        if(counter > 0){
            if(height == 1){                                
                System.out.println("move "+ height +" from " +
                                                source + " to " + destination);
            }else{  
                counter--;
                slide(counter, height - 1, source, destination, buffer);    
                System.out.println("move "+ hoehe +" from " +
                                                source + " to " + destination);
                counter--;
                slide(counter, height - 1, buffer, source, destination);    
            }
        }else{
            throw new Exception("stop here");
        }
    }
}

以下是实例: http ://ideone.com/xeN4x

我的问题是我得到了

move 1 from A to B
move 2 from A to C
move 1 from B to C
move 3 from A to B
move 1 from C to A
move 2 from C to B
java.lang.Exception: stop

作为输出.但是应该执行5次而不是6次幻灯片.有任何想法吗？

Answer 1

问题是你正在测试计数器是否大于或等于1,然后将其递减2.

counter--;
// ...
counter--;

这里的柜台可能会消极.你需要检查一下.