# 查找具有最大汉明距离的旋转 > 原文： [https://www.geeksforgeeks.org/find-a-rotation-with-maximum-hamming-distance/](https://www.geeksforgeeks.org/find-a-rotation-with-maximum-hamming-distance/) 给定 n 个元素的数组，请创建一个新数组，该数组是给定数组的旋转，并且两个数组之间的汉明距离最大。 [两个长度相等的数组或字符串之间的汉明距离](https://en.wikipedia.org/wiki/Hamming_distance)是相应字符（元素）不同的位置数。 **注意**：对于给定的输入，可以有多个输出。 例子： ``` Input : 1 4 1 Output : 2 Explanation: Maximum hamming distance = 2. We get this hamming distance with 4 1 1 or 1 1 4 input : N = 4 2 4 8 0 output : 4 Explanation: Maximum hamming distance = 4 We get this hamming distance with 4 8 0 2. All the places can be occupied by another digit. Other solutions can be 8 0 2 4, 4 0 2 8 etc. ``` 创建另一个数组，该数组的大小是原始数组的两倍，以使此新数组（复制数组）的元素只是原始数组的元素，并按相同顺序重复两次。 例如，如果原始数组为`1 4 1`，则副本数组为`1 4 1 1 4 1`。 现在，遍历副本数组并查找每次移位（或旋转）的汉明距离。 因此，我们检查`4 1 1, 1 1 4, 1 4 1`，选择汉明距离最大的输出。 下面是上述方法的实现： ## C++ ```cpp // C++ program to Find another array // such that the hamming distance  // from the original array is maximum #include <bits/stdc++.h> using namespace std; // Return the maximum hamming distance of a rotation int maxHamming(int arr[], int n) {     // arr[] to brr[] two times so that     // we can traverse through all rotations.     int brr[2 *n + 1];     for (int i = 0; i < n; i++)         brr[i] = arr[i];     for (int i = 0; i < n; i++)          brr[n+i] = arr[i];     // We know hamming distance with 0 rotation     // would be 0\.     int maxHam = 0;         // We try other rotations one by one and compute     // Hamming distance of every rotation     for (int i = 1; i < n; i++)     {         int currHam = 0;         for (int j = i, k=0; j < (i + n); j++,k++)              if (brr[j] != arr[k])                  currHam++;         // We can never get more than n.          if (currHam == n)             return n;         maxHam = max(maxHam, currHam);     }     return maxHam; } // driver program int main()  {     int arr[] = {2, 4, 6, 8};         int n = sizeof(arr)/sizeof(arr[0]);     cout << maxHamming(arr, n);         return 0; } ```