Java 快速排序的两种实现在时间上有很大差异
我有两个快速排序的实现,第一个使用中间值(fist,middle,last)作为轴心,第二个使用中间元素作为轴心 第一次实施:Java 快速排序的两种实现在时间上有很大差异,java,algorithm,quicksort,Java,Algorithm,Quicksort,我有两个快速排序的实现,第一个使用中间值(fist,middle,last)作为轴心,第二个使用中间元素作为轴心 第一次实施: public class quickMedian { public void sort(int array[]) // pre: array is full, all elements are non-null integers // post: the array is sorted in ascending order { quickSort(a
public class quickMedian {
public void sort(int array[])
// pre: array is full, all elements are non-null integers
// post: the array is sorted in ascending order
{
quickSort(array, 0, array.length - 1); // quicksort all the elements in the array
}
public void quickSort(int array[], int start, int end)
{
int i = start; // index of left-to-right scan
int k = end; // index of right-to-left scan
if (end - start >= 1) // check that there are at least two elements to sort
{
if (array[start+(end-start)/2]>array[end]){
swap(array,start+(end-start)/2, end);
}
if (array[start]>array[end]){
swap(array,start, end);
}
if (array[start+(end-start)/2]>array[start]){
swap(array, start+(end-start)/2, start);
}
int pivot = array[start]; // set the pivot as the first element in the partition
while (k > i) // while the scan indices from left and right have not met,
{
while (array[i] <= pivot && i <= end && k > i) // from the left, look for the first
i++; // element greater than the pivot
while (array[k] > pivot && k >= start && k >= i) // from the right, look for the first
k--; // element not greater than the pivot
if (k > i) // if the left seekindex is still smaller than
swap(array, i, k); // the right index, swap the corresponding elements
}
swap(array, start, k); // after the indices have crossed, swap the last element in // the left partition with the pivot
quickSort(array, start, k - 1); // quicksort the left partition
quickSort(array, k + 1, end); // quicksort the right partition
}
else // if there is only one element in the partition, do not do any sorting
{
return; // the array is sorted, so exit
}
}
public void swap(int array[], int index1, int index2)
// pre: array is full and index1, index2 < array.length
// post: the values at indices 1 and 2 have been swapped
{
int temp = array[index1]; // store the first value in a temp
array[index1] = array[index2]; // copy the value of the second into the first
array[index2] = temp; // copy the value of the temp into the second
}
}
public class quickSort {
private int array[];
private int length;
public void sort(int[] inputArr) {
if (inputArr == null || inputArr.length == 0) {
return;
}
this.array = inputArr;
length = inputArr.length;
quickSorter(0, length - 1);
}
private void quickSorter(int lowerIndex, int higherIndex) {
int i = lowerIndex;
int j = higherIndex;
// calculate pivot number, I am taking pivot as middle index number
int pivot = array[lowerIndex+(higherIndex-lowerIndex)/2];
// Divide into two arrays
while (i <= j) {
/**
* In each iteration, we will identify a number from left side which
* is greater then the pivot value, and also we will identify a number
* from right side which is less then the pivot value. Once the search
* is done, then we exchange both numbers.
*/
while (array[i] < pivot) {
i++;
}
while (array[j] > pivot) {
j--;
}
if (i <= j) {
exchangeNumbers(i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort() method recursively
if (lowerIndex < j)
quickSorter(lowerIndex, j);
if (i < higherIndex)
quickSorter(i, higherIndex);
}
private void exchangeNumbers(int i, int j) {
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
只需复制第二个解决方案并插入函数对3个元素进行排序:sort3(int array[],int idx1,int idx2,int idx3)到开头…你是说选择pivot?第一个算法对我来说很奇怪(但第二个很好)。只需编写sort3()函数并修改第二个解决方案,如下所示:int i=lowerIndex;int j=高指数;sort3(数组,lowerIndex,lowerIndex+(higherIndex-lowerIndex)/2,higherIndex);int pivot=数组[lowerIndex+(higherIndex-lowerIndex)/2];
public static void main(String[] args) {
File number = new File ("f.txt");
final int size = 10000000;
try{
// quickSort s = new quickSort();
quickMedian s = new quickMedian();
writeTofile(number, size);
int [] arr1 =readFromFile(number, size);
long a=System.currentTimeMillis();
s.sort(arr1);
long b=System.currentTimeMillis();
System.out.println("quickSort: "+(double)(b-a)/1000);
}catch (Exception ex){ex.printStackTrace();}
}