Perfect Shuffle algorithm

Perfect Shuffle algorithm Topic Details: There is an array of length 2n {a1,a2,a3,..., an,b1,b2,b3,..., bn}, if you wish to sort {a1,b1,a2,b2,...., an,bn}, consider a solution with no time complexity O (n), spatial complexity 0 (1). http://blog.csdn.net/starcuan/article/details/19079913

#include <iostream> #include <algorithm> #include <functional> using namespace std; /* Perfect Shuffle algorithm topic details: There is an array of length 2n {a1,a2,a3,..., an,b1,b2,b3,..., bn}, you want to sort {a1,b1,a2,b2,...., an,bn}, consider there is no time complexity O (n), The solution of Space complexity 0 (1). */#define N 1000//Time O (n), the space O (n) array subscript starting from 1//any of the element I, and finally to (2 * i)% (2 * N + 1) position void pefect_shuffle1 (int *a,        int n) {int n2 = n * 2, I, b[n];        for (i = 1; I <= n2; ++i) {b[(i * 2)% (N2 + 1)] = A[i];        } for (i = 1; I <= n2; ++i) {a[i] = B[i];          }}//Time O (nlogn) space O (1) array subscript starting from 1 void perfect_shuffle2 (int *a,int n) {int t,i;              if (n = = 1) {t = a[1];              A[1] = a[2];              A[2] = t;          Return          } int n2 = n * 2, n3 = N/2;              if (n% 2 = = 1) {//odd processing t = A[n];              for (i = n + 1; I <= n2; ++i) {a[i-1] = A[i];          }    A[N2] = t;          --n;              }//To this n is an even for (i = n3 + 1; I <= n; ++i) {t = A[i];              A[i] = a[i + n3];          A[i + n3] = t;          }//[1.. n/2] Perfect_shuffle2 (A, N3);      Perfect_shuffle2 (A + N, N3);  }/* First lap: 1, 2, 4, 8, 7, 5, 1--and 3, 6, 3: Original array: 1 2 3 4 5 6 7 8 Array of small Labels: 1 2 3 4 5 6 7 8 go first lap: 5 1 (3) 2 7 (6) 8 4 Go second lap: 5 1 6 2 7 3 8 4 */* Divine Conclusion: If 2*n= (3^k-1), the number of rings can be determined and their respective The starting position of the head for 2*n = (3^k-1) This length of the array, just a K-ring, and each lap head starting position is 1,3,9, ... 3^ (k-1).  *//array subscript starting from 1, the From is the head of the Circle, MoD is to modulo the number mod should be 2 * n + 1, time complexity O (circle length) void Cycle_leader (int *a,int from, int mod) {int         last = A[from],t,i;            for (i = from * 2 mod;i! = from; i = i * 2 mod) {t = A[i];            A[i] = last;           last = t;    } A[from] = last; }//Flip string time complexity O (to-from) void reverse (int *a,intFrom,int to) {int t;            for (; from < to; ++from,--to) {t = A[from];            A[from] = a[to];        A[to] = t;        }}//Loop right shift num bit time complexity O (n) void right_rotate (int *a,int num,int N) {reverse (A, 1, n-num);        Reverse (A, n-num + 1,n);    Reverse (A, 1, n);  }/* Perfect shuffle algorithm, the algorithm flow is: input array a[1..2 * N] Step 1 found 2 * m = 3^k-1 make 3^k <= 2 * N < 3^ (k + 1) Step 2 a[m     + 1..N + m] that part of the loop moves the M-bit step 3 to each i = 0,1,2..k-1,3^i is the head of the ring, do the cycle_leader algorithm, the array length is M, so to 2 * m + 1 modulo. Step 4 The back part of the array a[2 * m + 1:2 * n] continues to use this algorithm, which is equivalent to n reduced m.        *//Time O (n), Space O (1) void perfect_shuffle3 (int *a,int n) {int n2, M, I, k,t; for (; n > 1;)            {//step 1 n2 = n * 2;            for (k = 0, M = 1; n2/m >= 3; ++k, M *= 3);            M/= 2;                        2m = 3^k-1, 3^k <= 2n < 3^ (k + 1)//Step 2 right_rotate (A + M, m, N); // Step 3 for (i = 0, t = 1; i < K; ++i, T *= 3) {Cycle_leader (A, T, M * 2 + 1);            }//step 4 A + = M * 2;        N-= m;        }//n = 1 t = a[1];        A[1] = a[2];    A[2] = t;      } int main () {int arr[]={0,1,2,3,4,5,6,7,8};      Perfect_shuffle3 (arr,4);      For_each (arr,arr+9,[] (int i) {cout<<i<<endl;});  return 0;  }

　　

Perfect Shuffle algorithm