用C语言代码来编写含汉诺塔问题,利用堆栈来实现.求代码 脚本语言和编程语言的区别是什么？
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\u5982\u679c\u81ea\u5b66\uff0c\u53ea\u80fd\u517b\u6210\u597d\u7684\u5b66\u4e60\u4e60\u60ef\u6162\u6162\u5b66\uff0c\u65e5\u79ef\u6708\u7d2f\uff0c\u7ec8\u80fd\u5b66\u6709\u6240\u6210\u7684\uff01\u4e0d\u8fc7\u81ea\u5236\u529b\u4e0d\u597d\u7684\u4eba\u4e0d\u5efa\u8bae\u81ea\u5b66\u3002
\u5982\u679c\u8981\u4e0a\u57f9\u8bad\u673a\u6784\u5b66\uff0c\u5efa\u8bae\uff1a
\u8bfe\u524d\u9884\u4e60
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2.\u8bfe\u4e0a\u8ba4\u771f\u542c\u8bfe
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3.\u8bfe\u4e0b\u9879\u76ee\u590d\u4e60
\u8bfe\u4e0a\u8bb2\u8fc7\u7684\u4e1c\u897f\u4e0d\u80fd\u5149\u770b\u4f1a\u5c31\u591f\u4e86\uff0c\u4e00\u5b9a\u8981\u81ea\u5df1\u4eb2\u81ea\u4e0a\u624b\u53bb\u5b9e\u64cd\u4e00\u4e0b\u770b\u770b\uff0c\u53ea\u6709\u81ea\u5df1\u505a\u7684\u65f6\u5019\u624d\u77e5\u9053\u5230\u5e95\u5b66\u5230\u4ec0\u4e48\u7a0b\u5ea6\uff0c\u8fd8\u6709\u54ea\u91cc\u6ca1\u6709\u638c\u63e1\u3002
4.\u8bb0\u7b14\u8bb0
\u597d\u8bb0\u6027\u4e0d\u5982\u70c2\u7b14\u5934\uff0c\u5f53\u4f60\u9047\u5230\u89e3\u51b3\u4e0d\u4e86\u7684\u95ee\u9898\uff0c\u81ea\u5df1\u53ef\u4ee5\u627e\u4e00\u4e2a\u7b14\u8bb0\u672c\u8bb0\u5f55\u4e0b\u6765\uff0c\u8bfe\u4e0a\u8001\u5e08\u8bb2\u7684\u4e1c\u897f\u611f\u89c9\u4e0d\u662f\u5f88\u660e\u767d\u4e5f\u8981\u8bb0\u4e0b\u6765\uff0c\u6ca1\u4e8b\u7ffb\u770b\u4e00\u4e0b\u907f\u514d\u4ee5\u540e\u518d\u72af\u540c\u6837\u7684\u9519\u3002
5.\u4e0d\u7528\u614c\u5f20
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1\u3001\u811a\u672c\u8bed\u8a00\uff1a\u53c8\u88ab\u79f0\u4e3a\u6269\u5efa\u7684\u8bed\u8a00\uff0c\u6216\u8005\u52a8\u6001\u8bed\u8a00\uff0c\u662f\u4e00\u79cd\u7f16\u7a0b\u8bed\u8a00\uff0c\u7528\u6765\u63a7\u5236\u8f6f\u4ef6\u5e94\u7528\u7a0b\u5e8f\u3002
2\u3001\u7f16\u7a0b\u8bed\u8a00\uff1a\u4e00\u79cd\u8ba1\u7b97\u673a\u8bed\u8a00\u8ba9\u7a0b\u5e8f\u5458\u80fd\u591f\u51c6\u786e\u5730\u5b9a\u4e49\u8ba1\u7b97\u673a\u6240\u9700\u8981\u4f7f\u7528\u7684\u6570\u636e\uff0c\u5e76\u7cbe\u786e\u5730\u5b9a\u4e49\u5728\u4e0d\u540c\u60c5\u51b5\u4e0b\u6240\u5e94\u5f53\u91c7\u53d6\u7684\u884c\u52a8\u3002
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2\u3001\u7f16\u7a0b\u8bed\u8a00\uff1a\u7a0b\u5e8f\u5728\u8fd0\u884c\u65f6\u53ef\u4ee5\u6539\u53d8\u5176\u7ed3\u6784\uff0c\u5f00\u53d1\u5feb\u901f\uff0c\u63a5\u8fd1\u81ea\u7136\u8bed\u8a00\uff0c\u6613\u4e8e\u7406\u89e3\uff0c\u66f4\u65b9\u4fbf\u7684\u4ee3\u7801\u7ba1\u7406\u3002
\u4e09\u3001\u4f18\u7f3a\u70b9\u4e0d\u540c
1\u3001\u811a\u672c\u8bed\u8a00\uff1a\u811a\u672c\u8bed\u8a00\u6781\u5927\u5730\u7b80\u5316\u4e86\u201c\u5f00\u53d1\u3001\u90e8\u7f72\u3001\u6d4b\u8bd5\u548c\u8c03\u8bd5\u201d\u7684\u5468\u671f\u8fc7\u7a0b\u3002\u5927\u591a\u6570\u811a\u672c\u8bed\u8a00\u90fd\u80fd\u591f\u968f\u65f6\u90e8\u7f72\uff0c\u800c\u4e0d\u9700\u8981\u8017\u65f6\u7684\u7f16\u8bd1/\u6253\u5305\u8fc7\u7a0b\u3002
2\u3001\u7f16\u7a0b\u8bed\u8a00\uff1a\u4efb\u4f55\u7f16\u7a0b\u8bed\u8a00\u7684\u4f7f\u7528\u90fd\u9700\u8981\u786e\u4fdd\u7f16\u5199\u7684\u7a0b\u5e8f\u80fd\u591f\u6839\u636e\u5b9e\u9645\u9700\u8981\u53ca\u65f6\u8c03\u6574\u3002\u56e0\u6b64\uff0c\u5feb\u901f\u539f\u578b\u5f00\u53d1\u73af\u5883\u975e\u5e38\u91cd\u8981\uff0c\u5728\u8fd9\u79cd\u73af\u5883\u4e0b\uff0c\u52a8\u6001\u8bed\u8a00\u548c\u76f8\u5173\u5e93\u7684\u7ed3\u5408\u53ef\u4ee5\u5927\u5927\u589e\u5f3a\u5176\u4f18\u52bf\u3002
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\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1-\u7f16\u7a0b\u8bed\u8a00
对于汉诺塔问题，当只移动一个圆盘时，直接将圆盘从 A 针移动到 C 针。若移动的圆盘为 n(n>1)，则分成几步走：把 (n-1) 个圆盘从 A 针移动到 B 针（借助 C 针）；A 针上的最后一个圆盘移动到 C 针；B 针上的 (n-1) 个圆盘移动到 C 针（借助 A 针）。每做一遍，移动的圆盘少一个，逐次递减，最后当 n 为 1 时，完成整个移动过程。
因此，解决汉诺塔问题可设计一个递归函数，利用递归实现圆盘的整个移动过程，问题的解决过程是对实际操作的模拟。
程序代码
#include <stdio.h>
int main()
{
int hanoi(int,char,char,char);
int n,counter;
printf("Input the number of diskes：");
scanf("%d",&n);
printf("\n");
counter=hanoi(n,'A','B','C');
return 0;
}
int hanoi(int n,char x,char y,char z)
{
int move(char,int,char);
if(n==1)
move(x,1,z);
else
{
hanoi(n-1,x,z,y);
move(x,n,z);
hanoi(n-1,y,x,z);
}
return 0;
}
int move(char getone,int n,char putone)
{
static int k=1;
printf("%2d:%3d # %c---%c\n",k,n,getone,putone);
if(k++%3==0)
printf("\n");
return 0;
}
Leetcode-227. 用栈模拟汉诺塔问题
新视像
06-09
在经典的汉诺塔问题中，有 3 个塔和 N 个可用来堆砌成塔的不同大小的盘子。要求盘子必须按照从小到大的顺序从上往下堆 （如，任意一个盘子，其必须堆在比它大的盘子上面）。同时，你必须满足以下限制条件：
(1) 每次只能移动一个盘子。(2) 每个盘子从堆的顶部被移动后，只能置放于下一个堆中。(3) 每个盘子只能放在比它大的盘子上面。
请写一段程序，实现将第一个堆的盘子移动到最后一个堆中。
样例
样例1:
输入: 3
输出:
towers[0]: []
towers[1]: []
towers[2]: [2,1,0]
样例2:
输入: 10
输出:
towers[0]: []
towers[1]: []
towers[2]: [9,8,7,6,5,4,3,2,1,0]
public class Tower {
private Stack<Integer> disks;
/*
* @param i: An integer from 0 to 2
*/
public Tower(int i) {
// create three towers
disks = new Stack<Integer>();
}
/*
* @param d: An integer
* @return: nothing
*/
public void add(int d) {
// Add a disk into this tower
if (!disks.isEmpty() && disks.peek() <= d) {
System.out.println("Error placing disk " + d);
} else {
disks.push(d);
}
}
/*
* @param t: a tower
* @return: nothing
*/
public void moveTopTo(Tower t) {
// Move the top disk of this tower to the top of t.
if (t.disks.isEmpty() || (!disks.isEmpty() && t.disks.peek() >= disks.peek())) {
t.disks.push(disks.pop());
}
}
/*
* @param n: An integer
* @param destination: a tower
* @param buffer: a tower
* @return: nothing
*/
public void moveDisks(int n, Tower destination, Tower buffer) {
// Move n Disks from this tower to destination by buffer tower
if(n<=0){
return;
}else if(n==1){
moveTopTo(destination);
}else{
moveDisks(n - 1, buffer, destination); //将源塔前几个盘子都放到辅助塔中
moveDisks(1, destination, buffer); //此时源塔只剩下一个最大的盘子，将它放到目标塔中
buffer.moveDisks(n - 1, destination, this); //最后再将辅助塔的全部盘子（因为此时目标塔上已经有了一个盘子，故辅助塔的盘子数为n-1）都放到目标塔上
}
}
/*
* @return: Disks
*/
public Stack<Integer> getDisks() {
// write your code here
return disks;
}
}
/**
* Your Tower object will be instantiated and called as such:
* Tower[] towers = new Tower[3];
* for (int i = 0; i < 3; i++) towers[i] = new Tower(i);
* for (int i = n - 1; i >= 0; i--) towers[0].add(i);
* towers[0].moveDisks(n, towers[2], towers[1]);
* print towers[0], towers[1], towers[2]
*/
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