我有一个河内问题塔。我有三个堆栈,我想使用称为move()的方法以一定顺序递归移动元素。我应该在另一个从堆栈中添加和删除元素的类中使用方法pop()和push()。
我尝试实现此方法,但它仅适用于第一种基本情况,例如。如果n == 1。
public static void move(SpecialStack from,
SpecialStack to,
SpecialStack help,
int n) {
if (n == 1) {
to.push(from.pop());
}
else {
move(from, to, help, n-1);
to.push(from.pop());
move(help, to, from, n-1);
}
}
public static void main(String[] args) {
int size = 3;
SpecialStack from = new SpecialStack(size);
SpecialStack to = new SpecialStack();
SpecialStack help = new SpecialStack();
System.out.println("Start state");
System.out.println(" From: " + from);
System.out.println(" To: " + to);
System.out.println(" Help: " + help);
move(from, to, help, size);
System.out.println("End state");
System.out.println(" From: " + from);
System.out.println(" To: " + to);
System.out.println(" Help: " + help);
我从push()方法得到的错误是我自己的错误,“数字太高!”来自另一个类SpecialStack:
import java.util.*;
import java.util.ArrayList;
public class SpecialStack {
private ArrayList<Integer> specStack;
public SpecialStack() {
specStack = new ArrayList<Integer>();
}
public SpecialStack(int n) {
this.specStack = new ArrayList<Integer>(n);
int i;
for (i=0; i<n; i++) {
specStack.add(i, n-i);
}
}
public void push(int x) {
if (specStack.size() == 0) {
specStack.add(x);
}
else if (x > specStack.get(specStack.size() -1)) {
throw new RuntimeException("Number too high");
}
else {
specStack.add(x);
}
}
public int pop() {
if (specStack.size() == 0) {
throw new RuntimeException("Empty stack");
}
else {
int length = specStack.size() -1;
int topNumber = specStack.get(length);
specStack.remove(length);
return topNumber;
}
}
public String toString() {
String arrList = "[";
int i;
for (i = 0; i < specStack.size(); i++) {
if (i == specStack.size() -1) {
arrList = arrList + specStack.get(i);
}
else {
arrList = arrList + specStack.get(i) + ",";}
}
arrList = arrList + "]";
return arrList;
}
}
我不明白为什么会出现此错误。而且,move()方法正确吗?感谢您的建议!
答案 0 :(得分:0)
您的代码中有一个愚蠢的错误。在您的其他部分,当n > 1
时,您应该做
move(from, help, to, n-1);
代替
move(from, to, help, n-1);
请记住,对于大于1的任何事物,您都将“帮助”队列视为除一个钉子之外的所有钉子的目的地队列。完成此操作后,将最低钉子移到原始的“ to”队列,并考虑原始的“ help”为“ from”和原始的“ from”为“ help”来重复这些操作。希望能有所帮助。