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汉诺塔问题
古代有一个梵塔,塔内有三个座A、B、C,A座上有64个盘子
,盘子大小不等,大的在下,小的在上(如图)。有一个和尚
想把这64个盘子从A座移到B座,但每次只能允许移动一个盘子
,并且在移动过程中,3个座上的盘子始终保持大盘在下,小盘
在上。在移动过程中可以利用B座,要求输出移动的步骤。
汉诺塔问题递归解法
C++代码
1 //By LYLtim 2 //2015.2.8 3 4 #include <iostream> 5 6 using namespace std; 7 8 void hanoi(int n, char src, char mid, char dst) { 9 if (n == 1) 10 cout << src << "->" << dst << endl; 11 else { 12 hanoi(n-1, src, dst, mid); 13 cout << src << "->" << dst << endl; 14 hanoi(n-1, mid, src, dst); 15 } 16 } 17 18 int main() 19 { 20 int n; 21 cin >> n; 22 hanoi(n, ‘A‘, ‘B‘, ‘C‘); 23 24 }
以输入3盘子为例输出
A->C
A->B
A->C
C->B
B->C
B->A
A->C
Java代码
1 //By LYLtim 2 //2011.11.12 3 4 public class hanoi { 5 6 public static void main(String[] args) { 7 movdDisk(3, 1, 3, 2); 8 } 9 10 static void movdDisk(int n, int start, int end, int mid) { 11 if (n == 1) 12 System.out.println("move disk 1 from post " + start + " to post " + end ); 13 else { 14 movdDisk(n - 1, start, mid, end); 15 System.out.println("move disk " + n + " from post " + start + " to post " + end); 16 movdDisk(n - 1, mid, end, start); 17 } 18 } 19 }
汉诺塔问题非递归解法
1 //By LYLtim 2 //2015.2.8 3 4 #include <iostream> 5 #include <stack> 6 7 using namespace std; 8 9 struct Problem 10 { 11 int n; 12 char src, mid, dst; 13 Problem(){} 14 Problem(int n, char s, char m, char d) : n(n), src(s), mid(m), dst(d) {} 15 }; 16 17 int main() 18 { 19 int n; 20 cin >> n; 21 stack<Problem> stk; 22 Problem curPrb; 23 stk.push(Problem(n, ‘A‘, ‘B‘, ‘C‘)); 24 while (!stk.empty()) { 25 curPrb = stk.top(); 26 stk.pop(); 27 if (curPrb.n == 1) 28 cout << curPrb.src << "->" << curPrb.dst << endl; 29 else { 30 stk.push(Problem(curPrb.n-1, curPrb.mid, curPrb.src, curPrb.dst)); 31 cout << curPrb.src << "->" << curPrb.dst << endl; 32 stk.push(Problem(curPrb.n-1, curPrb.src, curPrb.dst, curPrb.mid)); 33 } 34 } 35 }
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原文地址:http://www.cnblogs.com/LYLtim/p/4280256.html