Question

我试图了解这种基数排序算法的工作原理。我是算法和比特的新手，所以这对我来说并不容易。到目前为止，我已将这些注释添加到我的代码中，以便更容易理解。我不确定我是否正确掌握了这个概念，所以如果有人能看到我的评论有任何问题/我不能正确理解，请帮助我：）

也有人能够向我解释这行代码： mask = 1＆lt;＆lt;位;

我的评论代码：

public static ArrayList<Integer> RadixSort(ArrayList<Integer> a)
    //This method implements the radix sort algorithm, taking an integer array as an input
    {
        ArrayList<Integer> array = CopyArray(a);
        //Created a new integer array called 'array' and set it to equal the array inputed to the method
        //This was done by copying the array entered to the method through the CopyArray method, then setting the results of the method to the new empty array

        Integer[] zerobucket = new Integer[a.size()];
        Integer[] onebucket = new Integer[a.size()];
        //Created two more integer arrays to act as buckets for the binary values
        //'zerobucket' will hold array elements where the ith bit is equal to 0
        //'onebucket' will hold array elements where the ith bit is equal to 1

        int i, bit;
        //Created two integer variables i & bit, these will be used within the for loops below
        //Both i & bit will be incremented to run the radix sort for every bit of the binary value, for every element in the array

        Integer element, mask;
        //Created an integer object called element, this will be used to retrieve the ith element of the unsorted array
        //Created an integer object called mask, this will be used to compare the bit values of each element

        for(bit=0; bit<8; ++bit)
        //Created a for loop to run for every bit of the binary value e.g.01000000
        //Change from 8 to 32 for whole integers - will run 4 times slower
        {
            int zcount = 0;
            int ocount = 0;
            //Created two integer variables to allow the 'zerobucket' and 'onebucket' arrays to be increment within the for loop below

            for(i=0; i<array.size(); ++i)
            //Created a nested for loop to run for every element of the unsorted array
            //This allows every bit for every binary value in the array
            {
                element = array.get(i);
                //Set the variable 'element' to equal the ith element in the array
                mask = 1 << bit;

                if ((element & mask) == 0)
                //If the selected bit of the binary value is equal to 0, run this code
                {
                    zerobucket[zcount++] = array.get(i);
                    //Set the next element of the 'zerobucket' array to equal the ith element of the unsorted array
                }
                else
                //Else if the selected but of the binary value is not equal to 0, run this code
                {
                    onebucket[ocount++] = array.get(i);
                    //Set the next element of the 'onebucket' array to equal the ith element of the unsorted array
                }
            }

            for(i=0; i<ocount; ++i)
            //Created a for loop to run for every element within the 'onebucket' array
            {
                array.set(i,onebucket[i]);
                //Appended the ith element of the 'onebucket' array to the ith position in the unsorted array
            }

            for(i=0; i<zcount; ++i)
            //Created a for loop to run for every element within the 'zerobucket' array
            {
                array.set(i+ocount,zerobucket[i]);
                //Appended the ith element of the 'zerobucket' array to the ith position in the unsorted array
            }
        }
        return(array);
        //Returned the sorted array to the method
    }

我没有写这个代码我试图理解

Answer 1

我会以相反的顺序回答你的问题......

mask = 1＆lt;＆lt;位;

由于优先规则，您可以将其写为

mask =（1＆lt;＆lt; bit）;

哪个更明显一点。取整数1（0x01），将其左移位位，并将其指定为掩码。因此，如果位为2，则掩码为00000010（跳过前导零）。如果位为4，则掩码为00001000.依此类推。

掩码的原因如下：

if（（element＆amp; mask）== 0）

用于识别作为位置位的位是1还是0。与位掩码进行AND运算的项将为零或非零，具体取决于位掩码中与位1相同位的位是0还是非零。

现在更复杂的问题。所讨论的算法是最低有效位基数排序，意味着基数排序，其中对排序值的传递从最小到最重要（或者在软件整数，从右到左）位。

以下伪代码描述了您的代码：

array = copy (a)
for every bit position from 0 to radix // radix is 8
    for each item in array
        if bit at bit position in item is 0
            put item in zeroes bucket
        else
            put item in ones bucket
    array = ordered concatenation of ones bucket and zeroes bucket
return array

那么为什么这样呢？您可以将此视为数组中项目的迭代加权排名。所有其他位相等，具有1位的项目将比具有0位的项目更大项目（对项目进行排名）。每次传球对最终位置变得更重要（对排名进行加权）。这些二进制排序的连续应用将导致更经常将1放置在1的桶中的项目。每当一个物品停留在1桶中时，如果其他物品没有，那么该物品会改善其在1桶中的相对位置，与未来通过的其他物品相比也可能具有1.在1桶中保持一致存在与位置的持续改进相关联，其逻辑极值将是数组中的最大值。

希望有所帮助。

理解基数排序算法

1 个答案: