如何在 Java 中实现中位数算法
How to implement the medians of medians algorithm in Java
我正在尝试在 Java 中实现中位数算法。该算法应确定一组数字的中位数。我尝试在维基百科上实现伪代码:
https://en.wikipedia.org/wiki/Median_of_medians
我遇到了缓冲区溢出,但不知道为什么。由于递归,我很难跟踪代码。
import java.util.Arrays;
public class MedianSelector {
private static final int CHUNK = 5;
public static void main(String[] args) {
int[] test = {9,8,7,6,5,4,3,2,1,0,13,11,10};
lowerMedian(test);
System.out.print(Arrays.toString(test));
}
/**
* Computes and retrieves the lower median of the given array of
* numbers using the Median algorithm presented in the lecture.
*
* @param input numbers.
* @return the lower median.
* @throw IllegalArgumentException if the array is {@code null} or empty.
*/
public static int lowerMedian(int[] numbers) {
if(numbers == null || numbers.length == 0) {
throw new IllegalArgumentException();
}
return numbers[select(numbers, 0, numbers.length - 1, (numbers.length - 1) / 2)];
}
private static int select(int[] numbers, int left, int right, int i) {
if(left == right) {
return left;
}
int pivotIndex = pivot(numbers, left, right);
pivotIndex = partition(numbers, left, right, pivotIndex, i);
if(i == pivotIndex) {
return i;
}else if(i < pivotIndex) {
return select(numbers, left, pivotIndex - 1, i);
}else {
return select(numbers, left, pivotIndex + 1, i);
}
}
private static int pivot(int numbers[], int left, int right) {
if(right - left < CHUNK) {
return partition5(numbers, left, right);
}
for(int i=left; i<=right; i=i+CHUNK) {
int subRight = i + (CHUNK-1);
if(subRight > right) {
subRight = right;
}
int medChunk = partition5(numbers, i, subRight);
int tmp = numbers[medChunk];
numbers[medChunk] = numbers[(int) (left + Math.floor((double) (i-left)/CHUNK))];
numbers[(int) (left + Math.floor((double) (i-left)/CHUNK))] = tmp;
}
int mid = (right - left) / 10 + left +1;
return select(numbers, left, (int) (left + Math.floor((right - left) / CHUNK)), mid);
}
private static int partition(int[] numbers, int left, int right, int idx, int k) {
int pivotVal = numbers[idx];
int storeIndex = left;
int storeIndexEq = 0;
int tmp = 0;
tmp = numbers[idx];
numbers[idx] = numbers[right];
numbers[right] = tmp;
for(int i=left; i<right; i++) {
if(numbers[i] < pivotVal) {
tmp = numbers[i];
numbers[i] = numbers[storeIndex];
numbers[storeIndex] = tmp;
storeIndex++;
}
}
storeIndexEq = storeIndex;
for(int i=storeIndex; i<right; i++) {
if(numbers[i] == pivotVal) {
tmp = numbers[i];
numbers[i] = numbers[storeIndexEq];
numbers[storeIndexEq] = tmp;
storeIndexEq++;
}
}
tmp = numbers[right];
numbers[right] = numbers[storeIndexEq];
numbers[storeIndexEq] = tmp;
if(k < storeIndex) {
return storeIndex;
}
if(k <= storeIndexEq) {
return k;
}
return storeIndexEq;
}
//Insertion sort
private static int partition5(int[] numbers, int left, int right) {
int i = left + 1;
int j = 0;
while(i<=right) {
j= i;
while(j>left && numbers[j-1] > numbers[j]) {
int tmp = numbers[j-1];
numbers[j-1] = numbers[j];
numbers[j] = tmp;
j=j-1;
}
i++;
}
return left + (right - left) / 2;
}
}
确认n(在伪代码中)或i(在我的代码中)代表中位数的位置?所以让我们假设我们的数组是 number = {9,8,7,6,5,4,3,2,1,0}。我会调用 select{numbers, 0, 9,4), 对吗?
我不懂pivot中mid的计算?为什么要除以 10?也许伪代码有错误?
感谢您的帮助。
编辑:事实证明,从迭代到递归的切换是一个转移注意力的问题。 OP 确定的实际问题出现在第二次递归 select
调用的参数中。
这一行:
return select(numbers, left, pivotIndex + 1, i);
应该是
return select(numbers, pivotIndex + 1, right, i);
我会在下面留下原来的答案,因为我不想显得比实际聪明。
我认为您可能误解了 select
方法的伪代码 - 它使用迭代而不是递归。
这是您当前的实现:
private static int select(int[] numbers, int left, int right, int i) {
if(left == right) {
return left;
}
int pivotIndex = pivot(numbers, left, right);
pivotIndex = partition(numbers, left, right, pivotIndex, i);
if(i == pivotIndex) {
return i;
}else if(i < pivotIndex) {
return select(numbers, left, pivotIndex - 1, i);
}else {
return select(numbers, left, pivotIndex + 1, i);
}
}
和伪代码
function select(list, left, right, n)
loop
if left = right then
return left
pivotIndex := pivot(list, left, right)
pivotIndex := partition(list, left, right, pivotIndex, n)
if n = pivotIndex then
return n
else if n < pivotIndex then
right := pivotIndex - 1
else
left := pivotIndex + 1
这通常使用 while
循环来实现:
private static int select(int[] numbers, int left, int right, int i) {
while(true)
{
if(left == right) {
return left;
}
int pivotIndex = pivot(numbers, left, right);
pivotIndex = partition(numbers, left, right, pivotIndex, i);
if(i == pivotIndex) {
return i;
}else if(i < pivotIndex) {
right = pivotIndex - 1;
}else {
left = pivotIndex + 1;
}
}
}
通过此更改,您的代码似乎可以正常工作,但显然您需要进行测试以确认。
int[] test = {9,8,7,6,5,4,3,2,1,0,13,11,10};
System.out.println("Lower Median: " + lowerMedian(test));
int[] check = test.clone();
Arrays.sort(check);
System.out.println(Arrays.toString(check));
输出:
Lower Median: 6
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13]
我正在尝试在 Java 中实现中位数算法。该算法应确定一组数字的中位数。我尝试在维基百科上实现伪代码:
https://en.wikipedia.org/wiki/Median_of_medians
我遇到了缓冲区溢出,但不知道为什么。由于递归,我很难跟踪代码。
import java.util.Arrays;
public class MedianSelector {
private static final int CHUNK = 5;
public static void main(String[] args) {
int[] test = {9,8,7,6,5,4,3,2,1,0,13,11,10};
lowerMedian(test);
System.out.print(Arrays.toString(test));
}
/**
* Computes and retrieves the lower median of the given array of
* numbers using the Median algorithm presented in the lecture.
*
* @param input numbers.
* @return the lower median.
* @throw IllegalArgumentException if the array is {@code null} or empty.
*/
public static int lowerMedian(int[] numbers) {
if(numbers == null || numbers.length == 0) {
throw new IllegalArgumentException();
}
return numbers[select(numbers, 0, numbers.length - 1, (numbers.length - 1) / 2)];
}
private static int select(int[] numbers, int left, int right, int i) {
if(left == right) {
return left;
}
int pivotIndex = pivot(numbers, left, right);
pivotIndex = partition(numbers, left, right, pivotIndex, i);
if(i == pivotIndex) {
return i;
}else if(i < pivotIndex) {
return select(numbers, left, pivotIndex - 1, i);
}else {
return select(numbers, left, pivotIndex + 1, i);
}
}
private static int pivot(int numbers[], int left, int right) {
if(right - left < CHUNK) {
return partition5(numbers, left, right);
}
for(int i=left; i<=right; i=i+CHUNK) {
int subRight = i + (CHUNK-1);
if(subRight > right) {
subRight = right;
}
int medChunk = partition5(numbers, i, subRight);
int tmp = numbers[medChunk];
numbers[medChunk] = numbers[(int) (left + Math.floor((double) (i-left)/CHUNK))];
numbers[(int) (left + Math.floor((double) (i-left)/CHUNK))] = tmp;
}
int mid = (right - left) / 10 + left +1;
return select(numbers, left, (int) (left + Math.floor((right - left) / CHUNK)), mid);
}
private static int partition(int[] numbers, int left, int right, int idx, int k) {
int pivotVal = numbers[idx];
int storeIndex = left;
int storeIndexEq = 0;
int tmp = 0;
tmp = numbers[idx];
numbers[idx] = numbers[right];
numbers[right] = tmp;
for(int i=left; i<right; i++) {
if(numbers[i] < pivotVal) {
tmp = numbers[i];
numbers[i] = numbers[storeIndex];
numbers[storeIndex] = tmp;
storeIndex++;
}
}
storeIndexEq = storeIndex;
for(int i=storeIndex; i<right; i++) {
if(numbers[i] == pivotVal) {
tmp = numbers[i];
numbers[i] = numbers[storeIndexEq];
numbers[storeIndexEq] = tmp;
storeIndexEq++;
}
}
tmp = numbers[right];
numbers[right] = numbers[storeIndexEq];
numbers[storeIndexEq] = tmp;
if(k < storeIndex) {
return storeIndex;
}
if(k <= storeIndexEq) {
return k;
}
return storeIndexEq;
}
//Insertion sort
private static int partition5(int[] numbers, int left, int right) {
int i = left + 1;
int j = 0;
while(i<=right) {
j= i;
while(j>left && numbers[j-1] > numbers[j]) {
int tmp = numbers[j-1];
numbers[j-1] = numbers[j];
numbers[j] = tmp;
j=j-1;
}
i++;
}
return left + (right - left) / 2;
}
}
确认n(在伪代码中)或i(在我的代码中)代表中位数的位置?所以让我们假设我们的数组是 number = {9,8,7,6,5,4,3,2,1,0}。我会调用 select{numbers, 0, 9,4), 对吗?
我不懂pivot中mid的计算?为什么要除以 10?也许伪代码有错误?
感谢您的帮助。
编辑:事实证明,从迭代到递归的切换是一个转移注意力的问题。 OP 确定的实际问题出现在第二次递归 select
调用的参数中。
这一行:
return select(numbers, left, pivotIndex + 1, i);
应该是
return select(numbers, pivotIndex + 1, right, i);
我会在下面留下原来的答案,因为我不想显得比实际聪明。
我认为您可能误解了 select
方法的伪代码 - 它使用迭代而不是递归。
这是您当前的实现:
private static int select(int[] numbers, int left, int right, int i) {
if(left == right) {
return left;
}
int pivotIndex = pivot(numbers, left, right);
pivotIndex = partition(numbers, left, right, pivotIndex, i);
if(i == pivotIndex) {
return i;
}else if(i < pivotIndex) {
return select(numbers, left, pivotIndex - 1, i);
}else {
return select(numbers, left, pivotIndex + 1, i);
}
}
和伪代码
function select(list, left, right, n)
loop
if left = right then
return left
pivotIndex := pivot(list, left, right)
pivotIndex := partition(list, left, right, pivotIndex, n)
if n = pivotIndex then
return n
else if n < pivotIndex then
right := pivotIndex - 1
else
left := pivotIndex + 1
这通常使用 while
循环来实现:
private static int select(int[] numbers, int left, int right, int i) {
while(true)
{
if(left == right) {
return left;
}
int pivotIndex = pivot(numbers, left, right);
pivotIndex = partition(numbers, left, right, pivotIndex, i);
if(i == pivotIndex) {
return i;
}else if(i < pivotIndex) {
right = pivotIndex - 1;
}else {
left = pivotIndex + 1;
}
}
}
通过此更改,您的代码似乎可以正常工作,但显然您需要进行测试以确认。
int[] test = {9,8,7,6,5,4,3,2,1,0,13,11,10};
System.out.println("Lower Median: " + lowerMedian(test));
int[] check = test.clone();
Arrays.sort(check);
System.out.println(Arrays.toString(check));
输出:
Lower Median: 6
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13]