完美洗牌算法的多种实现

       非常喜欢看“v_JULY_v”大神的博客，可以说受益颇多。大神博客首页：http://blog.csdn.net/v_july_v?viewmode=contents。

昨天看了“程序员编程艺术第三十四~三十五章：格子取数问题，完美洗牌算法”，稍微实现了下文章提到的几种关于洗牌的算法，记录在此。

关于“完美洗牌”的理论，我在这不多讲了，请看上面的链接，这里只给出我自己实现的代码。

1.位置置换perfect_shuffle1算法

#define  NUM 9char * pa[9] = {NULL,"a1", "a2", "a3", "a4","b1", "b2", "b3", "b4"};void display_strings(char * pa[], int n){int i;for(i = 1; i < n; ++i)cout << pa[i] <<" ";cout << endl;}void perfect_shuffle1(char * pa[], const int n){int i = 0;char * pTmp[NUM];for (i = 1; i < n; ++i){pTmp[2 * i % n] = pa[i];}for (i = 1; i < n; ++i){pa[i] = pTmp[i];}}int main(){display_strings(pa, 9);perfect_shuffle1(pa, 9);display_strings(pa, 9);return 0;}

实验结果：

2.分而治之perfect_shuffle2算法

#define  NUM 9char * pa[9] = {NULL,"a1", "a2", "a3", "a4","b1", "b2", "b3", "b4"};char * pb[11] = {NULL,"a1", "a2", "a3", "a4", "a5","b1", "b2", "b3", "b4", "b5"};void display_strings(char * pa[], int n){int i;for(i = 1; i < n; ++i)cout << pa[i] <<" ";cout << endl;}void perfect_shuffle2(char * pa[], int start, int end){int n = (end - start + 1) / 2;int i = 0, j = 0;char * tmp;if (n == 1){return;}if (n & 0x01){tmp = pa[n];for (i = n; i <= 2 * n - 2 ; i++){pa[i] = pa[i + 1];}pa[2 * n - 1] = tmp;n--;}for (i = n / 2 + 1, j = 1; i <= n; ++i, j++){tmp = pa[i + start - 1]; pa[i + start - 1] = pa[j + n + start - 1]; pa[j + n + start - 1] = tmp;}perfect_shuffle2(pa , 1, n);perfect_shuffle2(pa , n + 1, 2 * n);}int main(){cout << "---------------------------------\n";display_strings(pa, 9);perfect_shuffle2(pa, 1, 8);display_strings(pa, 9);cout << "---------------------------------\n";cout << "\n";cout << "---------------------------------\n";display_strings(pb, 11);perfect_shuffle2(pb, 1, 10);display_strings(pb, 11);cout << "---------------------------------\n";return 0;}

3.完美洗牌算法perfect_shuffle3---走圈法

1）递归实现，有点“分而治之”的味道

#define  NUM 9char * pa[9] = {NULL,"a1", "a2", "a3", "a4","b1", "b2", "b3", "b4"};char * pb[11] = {NULL,"a1", "a2", "a3", "a4", "a5","b1", "b2", "b3", "b4", "b5"};void display_strings(char * pa[], int n){int i;for(i = 1; i < n; ++i)cout << pa[i] <<" ";cout << endl;}void cycle(char * pa[], int start, int mod){char * last = pa[start], * tmp;int i;for (i = 2 * start % mod ; i != start; i = 2 * i % mod){tmp = pa[i];pa[i] = last;last = tmp;}pa[start] = tmp;}void rotate_right(char * a[], int from, int to){char * tmp;for (; from < to; ++from, --to){tmp = a[from]; a[from] = a[to]; a[to] = tmp;}}void perfect_shuffle3(char * a[], const int n){int m, k, i, order;if(n == 0)return;for (k = 0, order = 1; order <= 2 * n + 1; ){++k;order *= 3;}k--;m = (order / 3 - 1) / 2;rotate_right(a, m + 1, n);rotate_right(a, n + 1, n + m);rotate_right(a, m + 1, n + m);for (i = 0, order = 1; i < k; ++i){cycle(a, order, 2 * m + 1);order *= 3;}perfect_shuffle3(a + 2 * m, n - m);}int main(){cout << "---------------------------------\n";display_strings(pa, 9);perfect_shuffle3(pa, 4);display_strings(pa, 9);cout << "---------------------------------\n";cout << "\n";cout << "---------------------------------\n";display_strings(pb, 11);perfect_shuffle3(pb, 5);display_strings(pb, 11);cout << "---------------------------------\n";return 0;}

2）看了“v_JULY_v”的代码，感觉完全没必要用到递归，故修改了源代码，得到“迭代”版本。

#define  NUM 9char * pa[9] = {NULL,"a1", "a2", "a3", "a4","b1", "b2", "b3", "b4"};char * pb[11] = {NULL,"a1", "a2", "a3", "a4", "a5","b1", "b2", "b3", "b4", "b5"};void display_strings(char * pa[], int n){int i;for(i = 1; i < n; ++i)cout << pa[i] <<" ";cout << endl;}void cycle(char * pa[], int start, int mod){char * last = pa[start], * tmp;int i;for (i = 2 * start % mod ; i != start; i = 2 * i % mod){tmp = pa[i];pa[i] = last;last = tmp;}pa[start] = tmp;}void rotate_right(char * a[], int from, int to){char * tmp;for (; from < to; ++from, --to){tmp = a[from]; a[from] = a[to]; a[to] = tmp;}}void perfect_shuffle4(char * a[],  int n){int m, k, i, order;while (n > 0){for (k = 0, order = 1; order <= 2 * n + 1; ){++k;order *= 3;}k--;m = (order / 3 - 1) / 2;rotate_right(a, m + 1, n);rotate_right(a, n + 1, n + m);rotate_right(a, m + 1, n + m);for (i = 0, order = 1; i < k; ++i){cycle(a, order, 2 * m + 1);order *= 3;}a += 2 * m;n -= m;}}int main(){cout << "---------------------------------\n";display_strings(pa, 9);perfect_shuffle4(pa, 4);display_strings(pa, 9);cout << "---------------------------------\n";cout << "\n";cout << "---------------------------------\n";display_strings(pb, 11);perfect_shuffle4(pb, 5);display_strings(pb, 11);cout << "---------------------------------\n";return 0;}

结束！！！