用栈来求解hanoi塔问题

题目来源：程序员算法面试指南，牛客网出品；

第一种解法：递归法；

//解法一：用递归法;
    public static int hanoi(int n, String left, String mid, String right){
        if(n<1){
            return 0;
        }
        return process(n, left, mid, right, left, right);
    }
    
    public static int process(int n, String left, String mid, String right, String from, String to){
        //n=1时,结束条件
        if(n==1){
            //from或to有一个是mid,只需移动1次
            if(from.equals(mid) || to.equals(mid)){
                System.out.println("Move 1 from "+from+" to "+to);
                return 1;
            } else{
                //from和to都不是mid,需移动2次
                System.out.println("Move 1 from "+ from + " to "+mid);
                System.out.println("Move 1 from "+ mid + " to " + to);
                return 2;
            }
        } 
        // n > 1,
        // from 或 to 有一个是mid, 只需移动一次;
        if(from.equals(mid)||to.equals(mid)){
            String another = (from.equals(left)||to.equals(left))? right : left;
            int part1 = process(n-1, left, mid, right, from, another);
            int part2 = 1;
            System.out.println("Move " + n +" from "+from+" to "+ to);
            int part3 = process(n-1, left, mid, right, another, to);
            return part1 + part2 + part3;
        }else{
            // from、to都不是mid, 共需移动5步
            int part1 = process(n-1, left, mid, right, from, to);
            int part2 = 1;
            System.out.println("Move "+ n +" from "+from +" to "+mid);
            int part3 = process(n-1, left, mid, right, to, from);
            int part4 = 1;
            System.out.println("Move "+ n +" from "+mid +" to "+to);
            int part5 = process(n-1, left, mid, right, from, to);
            return part1 + part2 + part3 + part4 + part5;
        }
    }

第二种解法：用栈模拟

//第二种解法
    //用栈模拟
    public static int hanoi(int n){
        Stack<Integer> LS = new Stack<Integer>();
        Stack<Integer> MS = new Stack<Integer>();
        Stack<Integer> RS = new Stack<Integer>();
        
        for(int i=n; i>0; i--)
            LS.push(i);
        
        int counter = 1;    //记录移动次数
        int last_step = 1;    //记录前一次移动，1：从左边移到中间，2：从中间移到左边，3：从中间移动到右边，4：从右边移动到中间
        MS.push(LS.pop());    //第一次，必须是左边往中间移动
        System.out.println("Move 1 from left to mid");
        
        while(RS.size() != n){
            for(int this_step=1; this_step<5; this_step++){    //步骤循环，判断哪种是正确的移动步骤，每次只有一种正确移法；
                //当前步，不能跟上一步重复，也不能跟上一步相反（浪费移动步数）
                if(this_step == last_step ||
                        this_step + last_step == 3 || this_step + last_step == 7)
                    continue;
                switch (this_step){
                case 1: //从左边移动到中间
                    if(LS.empty()) break;    //如果左边栈是空的，则不能进行移动
                    if(MS.empty() || LS.peek() < MS.peek()){    //如果中间栈为空，则不需要判断后边的，否则MS.peek()会报错
                        MS.push(LS.pop());
                        System.out.println("Move " + MS.peek() + " from left to mid.");
                        counter++;
                        last_step = this_step;    //当前步在下一次移动时成为上一步
                    }
                    break;
                case 2:
                    if(MS.empty()) break;
                    if(LS.empty() || MS.peek() < LS.peek()){
                        LS.push(MS.pop());
                        System.out.println("Move " + LS.peek() + " from mid to left.");
                        counter ++;
                        last_step = this_step;
                    }
                    break;
                case 3:
                    if(MS.empty()) break;
                    if(RS.empty() || MS.peek() < RS.peek()){
                        RS.push(MS.pop());
                        System.out.println("Move "+ RS.peek() + " from mid to right.");
                        counter ++;
                        last_step = this_step;
                    }
                    break;
                case 4:
                    if(RS.empty()) break;
                    if(MS.empty() || MS.peek() > RS.peek()){
                        MS.push(RS.pop());
                        System.out.println("Move "+ MS.peek() + " from right to mid.");
                        counter++;
                        last_step = this_step;
                    }
                    break;
                default:
                    System.out.println("step choice error!");
                }
            }
        }
        return counter;
    }