我正在尝试通过递归来解决河内塔问题,同时使用堆栈。使用n#增加大小的磁盘,输出应如下所示。磁盘必须一次从一个Pillar1移动到Pillar3:
// assuming n = 3;
Pillar1: 3 2 1
Pillar2:
Pillar3:
// "\n"
Pillar1: 3 2
Pillar2:
Pillar3: 1
// "\n"
Pillar1: 3
Pillar2: 2
Pillar3: 1
// "\n"
Pillar1: 3
Pillar2: 2 1
Pillar3:
// "\n"
Pillar1:
Pillar2: 2 1
Pillar3: 3
// "\n"
Pillar1: 1
Pillar2: 2
Pillar3: 3
// "\n"
Pillar1: 1
Pillar2:
Pillar3: 3 2
// "\n"
Pillar1:
Pillar2:
Pillar3: 3 2 1
我的代码如下,我在输出磁盘时遇到困难> 1:
import java.util.*;
class TowersOfHanoiThree
{
public static Stack<Integer>[] tower = new Stack[4];
public static int temp;
public static void TowersOfHanoiThree(int numDisk)
{
//adding disk to stack
temp = numDisk;
tower = new Stack[4];
for(int a = 0; a <= 3; a++)
{
tower[a] = new Stack<Integer>();
}
for (int i = numDisk; i > 0; i--)
{
tower[1].push(numDisk);
show();
}
solver(numDisk, 1, 3, 2);
}
public static void show()
{
//System.out.println("Pillar1: ");
//System.out.println("Pillar2: ");
//System.out.println("Pillar3: ");
String Pillar1 = "Pillar1: ";
String Pillar2 = "Pillar2: ";
String Pillar3 = "Pillar3: ";
for(int x = temp -1 ; x >= 0 ; x--)
{
String emptStr1 = "";
String emptStr2 = "";
String emptStr3 = "";
try
{
emptStr1 = String.valueOf(tower[1].get(x));
}
catch(Exception e)
{
}
try
{
emptStr2 = String.valueOf(tower[2].get(x));
}
catch(Exception e)
{
}
try
{
emptStr3 = String.valueOf(tower[3].get(x));
}
catch(Exception e)
{
}
System.out.print(Pillar1+emptStr1+"\n");
System.out.print(Pillar2+emptStr2+"\n");
System.out.print(Pillar3+emptStr3+"\n");
System.out.print("\n");
}
}//end show
public static void solver(int numDisk, int start, int middle, int end)
{
if(numDisk > 0)
{
try
{
//sorting disks
solver(numDisk - 1, start, end, middle);
int dis = tower[start].pop(); //move disk top-most disk of start
tower[middle].push(dis);
show();
solver(numDisk - 1, middle, start, end);
}
catch(Exception e)
{
}
}
}
public static void main(String args[])
{
tower[1] = new Stack<Integer>();
tower[2] = new Stack<Integer>();
tower[3] = new Stack<Integer>();
TowersOfHanoiThree(2);
}
}
答案 0 :(得分:0)
使用以下实现应该可以提供所需的结果。我还对所做的修改提出了一些内联评论
public static void TowersOfHanoiThree(int numDisk)
{
//adding disk to stack
temp = numDisk;
tower = new Stack[4];
for(int a = 0; a <= 3; a++)
{
tower[a] = new Stack<Integer>();
}
for (int i = numDisk; i > 0; i--)
{
tower[1].push(i); // to show "1 2 3" i changed the value which was pushed in the stack
// from tower[1].push(numDisk) to tower[1].push(i)
}
show(); //In your example this method call was placed inside the for loop.
//Moving it outside will show the stacks only after the values are added
solver(numDisk, 1, 3, 2);
}
public static void show() {
// System.out.println("Pillar1: ");
// System.out.println("Pillar2: ");
// System.out.println("Pillar3: ");
String Pillar1 = "Pillar1: ";
String Pillar2 = "Pillar2: ";
String Pillar3 = "Pillar3: ";
String emptStr1 = "";
String emptStr2 = "";
String emptStr3 = "";
//the empStr variable are moved outside the loop because only after the
//loop has finished we know what values are in each pillar
for (int x = 0; x <= temp - 1; x++) {
try {
//here we just append the values from the pillar to string empStr1
emptStr1 += String.valueOf(tower[1].get(x)) + " ";
} catch (Exception e) {
}
try {
emptStr2 += String.valueOf(tower[2].get(x)) + " ";
} catch (Exception e) {
}
try {
emptStr3 += String.valueOf(tower[3].get(x)) + " ";
} catch (Exception e) {
}
}
//finally when the loop is done we have the results
System.out.print(Pillar1 + emptStr1 + "\n");
System.out.print(Pillar2 + emptStr2 + "\n");
System.out.print(Pillar3 + emptStr3 + "\n");
System.out.print("\n");
}// end show
答案 1 :(得分:0)
您需要解决一些问题。我会尽力解释它们并提供可能的解决方案。
首先,您始终将numDisks
推送到TowersOfHanoiThree(int)
方法的初始堆栈(支柱)上。
违规部分:
for (int i = numDisk; i > 0; i--) {
tower[1].push(numDisk); // should be tower[1].push(i);
show();
}
修正:
for (int i = numDisk; i > 0; i--) {
tower[1].push(i); // fixed
show();
}
其次,您的show()
方法由于多种原因而无法正确显示:
tower
)开始遍历数组temp - 1
中的每个堆栈,因为您要从底部到顶部显示磁盘。 emptStr1
循环之前声明您的字符串emptStr2
,emptStr3
,for
。目前,您每次迭代都会重置它们以清空字符串。emptStr1
,emptStr2
,emptStr3
,而不是通过重新分配来覆盖它们。emptStr1
,emptStr2
,emptStr3
。我建议完全重写show()
,例如:
public static void show() {
for (int i = 1; i < tower.length; i++) { // iterate over your array of stacks
String pillar = "Pillar " + i + ":"; // e.g. "Pillar 1: "
for (int disk : tower[i]) { // iterate over the stack located at index i
pillar += " " + disk; // concatenate the current disk to the String
}
System.out.println(pillar); // Display this pillar
}
System.out.println(); // Line break to seperate the show() calls
}
最后,solver(int, int, int, int)
中的第二次递归调用不正确。
违规行:
solver(numDisk - 1, middle, start, end);
这应该改为:
solver(numDisk - 1, end, middle, start);
另外,虽然不是失败的原因,但您的数组tower
的容量为4,但它应为3.您当前忽略索引0.我认为你是故意这样做的,因为你想把支柱显示为1,2和3?无论如何,这可能令人困惑。相反,当您显示支柱信息时,只需在索引中添加1即可。