用C ++实现的Radix Sort

我认为您的解决方案过于复杂。您可以使用输入中接收到的单个数组来实现基数，并且每个步骤中的存储桶都由一个索引数组来表示，这些索引标记了输入数组中每个存储桶的起始索引。

实际上，您甚至可以递归地执行此操作：

// Sort 'size' number of integers starting at 'input' according to the 'digit'th digit// For the parameter 'digit', 0 denotes the least significant digit and increases as significance doesvoid radixSort(int* input, int size, int digit){    if (size == 0)        return;    int[10] buckets;    // assuming decimal numbers    // Sort the array in place while keeping track of bucket starting indices.    // If bucket[i] is meant to be empty (no numbers with i at the specified digit),    // then let bucket[i+1] = bucket[i]    for (int i = 0; i < 10; ++i)    {        radixSort(input + buckets[i], buckets[i+1] - buckets[i], digit+1);    }}

当然

buckets[i+1] - buckets[i]

i

这样，您只需要调用就可以对

radixSort(testcases, sizeof(testcases) / sizeof(testcases[0]),0)