算法：完美洗牌算法：一个长度为2n的（整型）数组元素为 a1 a2 ... an b1 b2 ... bn

问题描述：一个长度为2n的数组元素为 a1 a2 ... an b1 b2 ... bn
要求： 用O(1)的空间代价, 在O(n)时间内把数组变成 b1 a1 b2 a2 b3 a3 ... bn an （分治法时间复杂度O(nlogn)）

采取置换的方法，当需要保证似乎循环无遗漏置换。需要用到数论原根等证明定理。

http://user.qzone.qq.com/414353346/blog/1243343118#!app=2&via=QZ.HashRefresh&pos=1243343118

算法论文：http://arxiv.org/pdf/0805.1598.pdf

自己的算法实现：

int static findMaxNear(int length){int k=3;while(k-1<length) k*=3;if(k-1==length)return k-1;elsereturn k/3 - 1;}void reverse(int a[],int start,int end){for(int i=start,j=end;i<j;i++,j--){int tmp=a[i];a[i]=a[j];a[j]=tmp;}}void rightmove(int a[],int start,int end,int steps){reverse(a,start,end);reverse(a,start,start+steps-1);reverse(a,start+steps,end);}void static shuffle_3k(int a[],int start,int length){int cyclestart=1; //3^0while(cyclestart<=length){ //length==3^k-1,cyclestart=1,3^1,...,3^(k-1)int tmp=a[start+cyclestart-1],nexttmp;int nextpos=2*cyclestart%(length+1); //nextpos为当前元素移动到的位置while(nextpos!=cyclestart){ //直到回到起点nexttmp=a[start+nextpos-1];a[start+nextpos-1]=tmp;tmp=nexttmp; nextpos=nextpos*2%(length+1);}a[start+cyclestart-1]=tmp;cyclestart*=3;}}void perfect_shuffle(int a[],int start,int length){int startpos=start,end=start+length-1;while(startpos-start<length){int cur_length=end-startpos+1;int m=findMaxNear(cur_length);//i=3^k-1<=Nif(m<cur_length) //还有剩余序列,则需要循环右移m/2个单位rightmove(a,startpos+m/2,startpos+cur_length/2+m/2-1,m/2);shuffle_3k(a,startpos,m);//分别以1,3,9,...,3^k为首元素开始循环置换startpos+=m;}}#define N 100void main(){int data[N]={0};int i=0;int n;printf("please input the number of data you wanna to test(should less than 100):\n");scanf("%d",&n);if(n&1){printf("sorry,the number should be even ");return;}for(i=0;i<n;i++)data[i]=i+1;perfect_shuffle(data,0,n); //in perfect_shuffle 首位均改变for(i=0;i<n;i++)printf("%d   ",data[i]);}

输入:10

即：  1 2 3 4 5 6 7 8 9 10

输出: 6 1 7 2 8 3 9 4 10 5