用棧求解漢諾塔問題

　　

　　漢諾塔問題比較經典，現在修改一下漢諾塔游戲的規則：規定不能直接從左移動到右，也不能直接從右移動到左，必須經過中柱，求解當塔有N層的時候，打印最優移動過程和最優移動總步數。

　　可以采用兩個方法解決。方法一采用遞歸的方法；方法二采用非遞歸的方法，用棧來模擬漢諾塔的三個塔。

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package com.test;import java.util.Stack;/*** Created by Demrystv.*/ public class SolveHanoiWithStack {//遞歸的方法，左->中，右->中，中->左，中->右，都是需要三個步驟，左->右，右->左，都是需要五個步驟public int hanoiProblem1(int num, String left, String mid, String right){if (num < 1){return 0;}return process(num, left, mid, right, left, right);}private 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 " + another);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;}}//非遞歸方法用到的枚舉類,因為不能直接從左到右，所以就只有四個動作public enum Action{No, LToM, MToL, MToR, RToM}//非遞歸的方法，用棧來模擬整個過程//核心思想：// 需要遵守兩個原則，// 一個是不能違反小壓大的原則，from棧彈出的元素num想壓入到to棧中，那么num的值必須小于等于當前to棧的棧頂// 另一個是相鄰不可逆原則，即不能上一步是左到中，下一步是中到左，不滿足最小步數// 由這兩個原則可以推出非遞歸方法的兩個結論：// 1.游戲的第一個動作一定是 左 到 中// 2.四個動作中只有一個動作和相鄰不可逆原則，其余三個一定都會違反，可以證明// 因此，每一步只有一個動作達標，那么只要每一步都根據這兩個原則考察所有的動作即可，哪個滿足就進行哪個動作，按照順序走下來即可。public 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 += fStackToStack(record, Action.MToL, Action.LToM, lS, mS, left, mid);step += fStackToStack(record, Action.LToM, Action.MToL, mS, lS, mid, left);step += fStackToStack(record, Action.RToM, Action.MToR, mS, rS, mid, right);step += fStackToStack(record, Action.MToR, Action.RToM, rS, mS, right, mid);}return step;}private int fStackToStack(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;} }

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轉載于:https://www.cnblogs.com/Demrystv/p/9292776.html

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