用栈来求解限制后的汉诺塔问题

本文主要是介绍用栈来求解限制后的汉诺塔问题，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

用栈来求解限制后的汉诺塔问题(限制不能从最左侧的塔直接移动到最右侧，也不能从最右侧直接移动到最左侧，而是必须经过中间，求当塔有N层的时候，打印最优移动过程和最优移动总步数)

import java.util.Stack;//用栈来求解限制后的汉诺塔问题(限制不能从最左侧的塔直接移动到最右侧，也不能从最右侧直接移动到最左侧，而是必须经过中间，求当塔有N层的时候，打印最优移动过程和最优移动总步数)
public class HanoiStack {//递归的方法public static int hanoiProblem1(int num, String left, String mid,String right) {if (num < 1) {return 0;}return process(num, left, mid, right, left, right);}public static int process(int num, String left, String mid, String right,String from, String to) {if (num == 1) {if (from.equals(mid) || to.equals(mid)) {System.out.println("Move 1 from " + from + " to " + to);return 1;} else {System.out.println("Move 1 from " + from + " to " + mid);System.out.println("Move 1 from " + mid + " to " + to);return 2;}}if (from.equals(mid) || to.equals(mid)) {String another = (from.equals(left) || to.equals(left)) ? right : left;int part1 = process(num - 1, left, mid, right, from, another);int part2 = 1;System.out.println("Move " + num + " from " + from + " to " + to);int part3 = process(num - 1, left, mid, right, another, to);return part1 + part2 + part3;} else {int part1 = process(num - 1, left, mid, right, from, to);int part2 = 1;System.out.println("Move " + num + " from " + from + " to " + mid);int part3 = process(num - 1, left, mid, right, to, from);int part4 = 1;System.out.println("Move " + num + " from " + mid + " to " + to);int part5 = process(num - 1, left, mid, right, from, to);return part1 + part2 + part3 + part4 + part5;}}
//=====================================================================================//枚举所有的操作public static enum Action {No, LToM, MToL, MToR, RToM}//非递归的方法，用栈来模拟汉诺塔的三个塔public static int hanoiProblem2(int num, String left, String mid, String right) {Stack<Integer> lS = new Stack<Integer>();Stack<Integer> mS = new Stack<Integer>();Stack<Integer> rS = new Stack<Integer>();lS.push(Integer.MAX_VALUE);mS.push(Integer.MAX_VALUE);rS.push(Integer.MAX_VALUE);for (int i = num; i > 0; i--) {lS.push(i);}Action[] record = { Action.No };int step = 0;while (rS.size() != num + 1) {step += fStackTotStack(record, Action.MToL, Action.LToM, lS, mS, left, mid);step += fStackTotStack(record, Action.LToM, Action.MToL, mS, lS, mid, left);step += fStackTotStack(record, Action.RToM, Action.MToR, mS, rS, mid, right);step += fStackTotStack(record, Action.MToR, Action.RToM, rS, mS, right, mid);}return step;}public static int fStackTotStack(Action[] record, Action preNoAct,Action nowAct, Stack<Integer> fStack, Stack<Integer> tStack,String from, String to) {if (record[0] != preNoAct && fStack.peek() < tStack.peek()) {tStack.push(fStack.pop());System.out.println("Move " + tStack.peek() + " from " + from + " to " + to);record[0] = nowAct;return 1;}return 0;}public static void main(String[] args) {int num = 2;   //汉诺塔的层数// solution 1(递归的方法)int steps1 = hanoiProblem1(num, "left", "mid", "right");System.out.println("It will move " + steps1 + " steps.");System.out.println("===================================");// solution 2(非递归的方法)int steps2 = hanoiProblem2(num, "left", "mid", "right");System.out.println("It will move " + steps2 + " steps.");System.out.println("===================================");}}

这篇关于用栈来求解限制后的汉诺塔问题的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---