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问题形容
给定三根柱子 A,B,C,柱子 A 上依照大小程序放着 n 个大小不同的盘子,最上面的最大,最下面的最小,输出盘子的个数,当初要把柱子 A 上的盘子全副挪动到 C 上。
问:
① 起码要挪动多少次,输入每一步挪动的形式。
② 输出开始计数的步骤和完结技术的步骤,统计在此期间每一个步骤呈现的次数。
代码
package Ring1270.pra.java01;
import java.sql.SQLOutput;
import java.util.Scanner;
/**
* https://www.lanqiao.cn/courses/2786/learning/?id=67029 */public class F_TowerOfHanoi {
//Constants
static int STEP = 0;
static int AB = 0;
static int AC = 0;
static int BC = 0;
static int BA = 0;
static int CA = 0;
static int CB = 0;
//Method of count
public static void count(String s) {if ("A-->B".equals(s)) AB++;
if ("A-->C".equals(s)) AC++;
if ("B-->A".equals(s)) BA++;
if ("B-->C".equals(s)) BC++;
if ("C-->A".equals(s)) CA++;
if ("C-->B".equals(s)) CB++;
}
//Method of output
public static void prin() {System.out.println("A-->B :" + AB);
System.out.println("A-->C :" + AC);
System.out.println("B-->A :" + BA);
System.out.println("B-->C :" + BC);
System.out.println("C-->A :" + CA);
System.out.println("C-->B :" + CB);
}
//Method of tower's moving
public static void move(int n, String from, String pass, String to, int x, int y) {if (n == 1) {System.out.println("Step" + ++STEP + ":" + "第 1 个盘子从" + from + "挪动到" + to + ",记为:" + from + "-->" + to);
if (STEP >= x) {
String s = from + "-->" + to;
count(s);
}
} else {
// 当有两个盘子的时候, 这里相当于把第一个挪动到 B 柱子上
move(n - 1, from, to, pass, x, y);
System.out.println("Step" + ++STEP + ":" + "将" + n + "个盘子从" + from + "挪动到" + to + ",记为:" + from + "-->" + to);
if (STEP >= x) {
String s = from + "-->" + to;
count(s);
}
// 而后把第二个盘子挪动,相当于把两头挪动到 C 下面去
move(n - 1, pass, from, to, x, y);
}
}
//Method of main
public static void main(String[] args) throws Exception {Scanner scanner = new Scanner(System.in);
System.out.println("Input the number of work's time:");
int n = scanner.nextInt();
System.out.println("Input the step of start:");
int x = scanner.nextInt();
System.out.println("Input the step of end:");
int y = scanner.nextInt();
if (x <= 0 || x > Math.pow(2, n) - 1 || x >= y) {throw new Exception("The start number is error , Runing is over......");
}
if (y <= 0 || y > Math.pow(2, n) - 1 || y <= x) {throw new Exception("The end number is error, Runing is over......");
}
int p = 0;
move(n, "A", "B", "C", x, y);
prin();}
}
运行截图
正文完