Perfect shuffling & shuffling

Perfect shuffling problem, given an array A1, A2, A3 ,... an, B1, B2, b3 .. BN, set it to B1, A1, B2, A2 ,... BN,.

O (n) algorithm, space of O (n.

For the first n numbers, the ing is f (I) = 2 * I + 1, 0 <= I <n/2; for example, 0-> 1, 1-> 3

For the next n numbers, map to f (I) = 2 (I-n/2), n/2 <= I <n; for example, n/2-> 0, n/2 + 1-> 2... and f (I) = 2 (I-n/2) = 2 * I-n = 2 * I + 1-(n + 1) = (2 * I + 1) % (n + 1 ).

Unified, ing to f (I) = (2 * I + 1) % (n + 1 ).

 1 void perfectShuffle1(int arr[], int n) { 2     int* tmp = new int[n]; 3     for (int i = 0; i < n; ++i) { 4         tmp[((i << 1) + 1) % (n + 1)] = arr[i]; 5     } 6     for (int i = 0; i < n; ++i) { 7         arr[i] = tmp[i]; 8     } 9     delete[] tmp;10 }

Division and control, O (nlgn) algorithm, and O (lgn) space.

Line 4-11 mainly deals with the odd half of the array.

When N/4! When the value is 0, the half of the array is an odd number. In this case, move the number of [n/2, n) to the left and put the number n/2-1 (the last number in the first half) to the end, the last two numbers are arranged. The problem is converted into an even half of the array.

When half of the array is an even number, you only need to swap the First Half of the array and the first half of the array to achieve the purpose of grouping.

 1 void perfectShuffle2(int arr[], int n) { 2     if (n % 2 != 0) return; 3     if (n <= 1) return; 4     if (n % 4 != 0) { 5         int tmp = arr[n / 2 - 1]; 6         for (int i = n / 2; i < n; ++i) { 7             arr[i - 1] = arr[i]; 8         } 9         arr[n - 1] = tmp;10         n -= 2;11     }12     for (int i = 0; i < n / 4; ++i) {13         swap(arr[n / 4 + i], arr[n / 2 + i]);14     }15     perfectShuffle2(arr, n / 2);16     perfectShuffle2(arr + n / 2, n / 2);17 }

Main reference from: http://blog.csdn.net/caopengcs/article/details/10176093, which mentioned the third solution did not go to see, I feel that the interview will not take the exam first effort to understand.

Shuffling is simple. In fact, it is equivalent to randomly selecting M numbers from the array. See the previous blog post. But here M = n.

1 void shuffle(int arr[], int n) {2     srand(time(NULL));3     for (int i = 0; i < n; ++i) {4         swap(arr[i], arr[i + rand() % (n - i)]);5     }6 }

This is the same principle as the fisheryates shuffling algorithm mentioned on the Internet.

Perfect shuffling & shuffling