《Cracking the Coding Interview》——第3章：栈和队列——题目4

2014-03-18 05:28

题目：你肯定听过汉诺威塔的故事：三个柱子和N个从小到大的盘子。既然每次你只能移动放在顶上的盘子，这不就是栈操作吗？所以，请用三个栈来模拟N级汉诺威塔的玩法。放心，N不会很大的。

解法：递归着玩儿吧，还挺容易写的。要是迭代，我估计够呛。

代码：

// 3.4 Implement Hanoi Tower with three stacks.
#include <cstdio>
#include <stack>
using namespace std;

class Solution {
public:
    void initHanoiTower(int n) {
        int i;
        
        clearHanoiTower();
        
        for (i = n; i >= 1; --i) {
            s[0].push(i);
        }
    }
    
    void moveHanoiTower(int n, int from, int to) {
        if (from == to) {
            return;
        }
        
        if ((int)s[from].size() < n) {
            return;
        }
        
        if (n == 1) {
            s[to].push(s[from].top());
            s[from].pop();
            return;
        }
        
        int i;
        for (i = 0; i < 3; ++i) {
            b[i] = false;
        }
        b[from] = true;
        b[to] = true;
        int other;
        for (i = 0; i < 3; ++i) {
            if (!b[i]) {
                other = i;
                break;
            }
        }

        moveHanoiTower(n - 1, from, other);
        moveHanoiTower(1, from, to);
        moveHanoiTower(n - 1, other, to);
    }
    
    void clearHanoiTower() {
        int i;
        
        for (i = 0; i < 3; ++i) {
            while (!s[i].empty()) {
                s[i].pop();
            }
        }
    }
    
    void printHanoiTower() {
        stack<int> ss[3];
        int i;
        
        for (i = 0; i < 3; ++i) {
            ss[i] = s[i];
            printf("Tower %d:", i + 1);
            while (!ss[i].empty()) {
                printf(" %d", ss[i].top());
                ss[i].pop();
            }
            printf("
");
        }
    }
private:
    stack<int> s[3];
    int b[3];
};

int main()
{
    int n;
    Solution sol;
    
    while (scanf("%d", &n) == 1 && n > 0) {
        sol.initHanoiTower(n);
        
        printf("At the beginning:
");
        sol.printHanoiTower();
        printf("
");

        sol.moveHanoiTower(n, 0, 2);
        
        printf("At the end:
");
        sol.printHanoiTower();
        printf("
");
    }
    
    return 0;
}