首先,我想从我的方法背后的推理开始。我的想法是,我要么喂食偶数或奇数的数组。这个数组将在二维数组中被分解为原始数组的长度,每个索引都有一个大小为1的数组。然后,我将继续合并索引。这适用于长度为2,4,8,16的长度。现在,我不得不编辑我的方法,因为3,5,6,7不工作。现在,我的问题是,即使他们工作,并非所有案件的工作。例如,25的长度不能正确返回。下面是我的代码:有一种工作合并排序,但不完全,有人可以告诉我我做错了什么?
/**
* A method to perform a merge sort
* @param array The array being fed in
*/
public static int[] mergeSort(int[] array)
{
if (array.length == 1)
{
return array;
}
int size = array.length;
int[][] miniArrayList = new int[size][1];
for (int index = 0; index < array.length; index++)
{
miniArrayList[index][0] = array[index];
}
while (miniArrayList.length > 1)
{
if (miniArrayList.length % 2 == 0)
{
miniArrayList = mergeEven(miniArrayList);
}
else
{
miniArrayList[0] = mergeOdd(miniArrayList);
}
}
return miniArrayList[0];
}
我对上述方法的想法是,我喂的,它打破它一个数组,然后一点一点地融合了阵,直到我有大小为1的数组排序。上述方法调用以下方法:
private static int[][] mergeEven(int[][] array)
{
int[][] tempSortList = new int[array.length/2][];
int tempIndex = 0;
for (int index = 0; index < array.length; index += 2)
{
tempSortList[tempIndex] = merge(array[index], array[index + 1]);
if (tempIndex != tempSortList.length)
{
tempIndex++;
}
}
array = tempSortList;
return array;
}
private static int[] mergeOdd(int[][] array)
{
/**
* The concept is to call the even merge method on the even part
* of the list and then once I get to one array, I merge that array
* with the extra array.
*/
int[][] localArray = new int[array.length - 1][1];
int[][] extra = new int[1][1];
for (int index = 0; index < localArray.length; index++)
{
localArray[index][0] = array[index][0];
}
extra[0][0] = array[array.length - 1][0];
int[][] tempSortList = new int[localArray.length/2][];
int tempIndex = 0;
for (int index = 0; index < localArray.length; index += 2)
{
tempSortList[tempIndex] = merge(localArray[index], localArray[index + 1]);
if (tempIndex != tempSortList.length)
{
tempIndex++;
}
}
localArray = tempSortList;
return localArray[0] = merge(localArray[0], extra[0]);
}
这是很自我解释,但以防万一。上面的方法排序不同,因为数组是奇数还是偶数。最后,这些方法调用实际的合并方法(工作我认为):
/**
* A merge method to merge smaller arrays to bigger
* arrays
* @param arrayOne The first array being fed in
* @param arrayTwo The second array being fed in
* @return A new integer array which is the sum of the
* length of the two arrays
*/
private static int[] merge(int[] arrayOne, int[] arrayTwo)
{
/*
* To make a proper method that deals with odd array sizes
* look into seeing if it odd, only merging length of the array -1
* and then once that is all merged merge that array with the last part of the array
*/
//Creating the size of the new subarray
int[] mergedArray;
if (arrayOne.length % 2 == 0 && arrayTwo.length % 2 == 0)
{
int size = arrayOne.length;
int doubleSize = 2 * size;
mergedArray = new int[doubleSize];
//Positions of each array
int posOne = 0;
int posTwo = 0;
int mergeIndex = 0;
while (posOne < size && posTwo < size)
{
if (arrayOne[posOne] < arrayTwo[posTwo])
{
mergedArray[mergeIndex] = arrayOne[posOne];
mergeIndex++;
posOne++;
}
else
{
mergedArray[mergeIndex] = arrayTwo[posTwo];
mergeIndex++;
posTwo++;
}
}
if (posOne == size)
{
for (int rest = posTwo; rest < size; rest++)
{
mergedArray[mergeIndex] = arrayTwo[rest];
mergeIndex++;
}
}
else
{
for (int rest = posOne; rest < size; rest++)
{
mergedArray[mergeIndex] = arrayOne[rest];
mergeIndex++;
}
}
}
else
{
int arrayOneSize = arrayOne.length;
int arrayTwoSize = arrayTwo.length;
int newArraySize = arrayOneSize + arrayTwoSize;
mergedArray = new int[newArraySize];
//Position in each array
int posOne = 0;
int posTwo = 0;
int mergeIndex = 0;
while (posOne < arrayOneSize && posTwo < arrayTwoSize)
{
if (arrayOne[posOne] < arrayTwo[posTwo])
{
mergedArray[mergeIndex] = arrayOne[posOne];
mergeIndex++;
posOne++;
}
else
{
mergedArray[mergeIndex] = arrayTwo[posTwo];
mergeIndex++;
posTwo++;
}
}
if (posOne == arrayOneSize)
{
mergedArray[mergeIndex] = arrayTwo[posTwo];
mergeIndex++;
}
else
{
for (int rest = posOne; rest < arrayOneSize; rest++)
{
mergedArray[mergeIndex] = arrayOne[rest];
mergeIndex++;
}
}
}
return mergedArray;
}
所以我的问题是,它的任何数组的大小我正确送入中,它只能在一定的尺寸。我已经阅读了一些文章,但是我的实现与我所见过的大多数不同,这就是我在这里发布的原因。我不想只是复制一个工作,我想了解我做错了什么,所以我可以修复它。预先感谢帮助家伙!
为了弄清楚什么是错误的,你需要找出代码中错误的地方。您可以使用调试器或打印调试输出来查看正在发生的情况,然后精确定位发生的情况与应发生的情况偏离。 – 2013-04-22 13:32:14