#include <iostream>
using namespace std;
//汉诺塔问题
void hano(int n, char a, char b, char c);
int main()
{
hano(3, 'A', 'B', 'C');
exit(0);
}
void hano(int n, char from, char to, char temp)
{
if (n == 1) {
cout << from << " => " << to << endl;
} else {
hano(n-1, from, temp, to);
hano(1, from, to, temp);
hano(n-1, temp, to, from);
}
}
//盘子带编号:
void move(int top, int n, char a, char b, char c);
int main()
{
move(1, 3, 'A', 'B', 'C');
exit(0);
}
void move(int top, int n, char from, char to, char temp)
{
if (n == 1) {
cout << top << ": " << from << " => " << to << endl;
} else {
hano(top, n-1, from, temp, to);
hano(top+n-1, 1, from, to, temp);
hano(top, n-1, temp, to, from);
}
}
本文介绍了经典的汉诺塔问题及其递归解决方案,并在此基础上引入了盘子带编号的实现方式,详细阐述了如何通过递归函数解决实际问题。
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