[英]Java Recursive MergeSort for ArrayLists
我的mergesort函數一直存在問題,因為我無法在將其輸入程序時對一系列整數或字符串進行排序。 我有一個調用項目的外部類,但它根本不對數字/字符串進行排序。 下面兩種方法,我不知道問題出在哪里。 數字是隨機輸入的。
碼:
/**
* Takes in entire vector, but will merge the following sections together:
* Left sublist from a[first]..a[mid], right sublist from a[mid+1]..a[last].
* Precondition: each sublist is already in ascending order
*
* @param a
* reference to an array of integers to be sorted
* @param first
* starting index of range of values to be sorted
* @param mid
* midpoint index of range of values to be sorted
* @param last
* last index of range of values to be sorted
*/
private void merge(ArrayList<Comparable> a, int first, int mid, int last) {
int x;
int i;
ArrayList<Comparable> left = new ArrayList<Comparable>();
ArrayList<Comparable> right = new ArrayList<Comparable>();
mergeSort(a,first,mid);
for(i = 0; i < a.size() - mid; i++){
left.add(i,a.get(i));
a.remove(i);
}
mergeSort(a,mid,last);
for (x = mid; x < a.size(); x++) {
right.add(x,a.get(x));
a.remove(x);
}
if ((left.get(i).compareTo(right.get(x))) > 0) {
i++;
a.add(i);
} else if (i < x) {
x++;
a.add(x);
}
System.out.println();
System.out.println("Merge");
System.out.println();
}
/**
* Recursive mergesort of an array of integers
*
* @param a
* reference to an array of integers to be sorted
* @param first
* starting index of range of values to be sorted
* @param last
* ending index of range of values to be sorted
*/
public void mergeSort(ArrayList<Comparable> a, int first, int last) {
int mid = (first + last)/2;
if(first == last){
}else if(last - first == 1){
merge(a,first, mid ,last);
}else{
last = mid;
}
}
我有一個調用項目的外部類,但它根本不對數字/字符串進行排序。 下面兩種方法,我不知道問題出在哪里。
第一個問題是,如果使用first = 0
和last = a.size()
調用mergeSort
方法, last = a.size()
會對任何內容進行排序,因為如果last-first == 1
則只調用merge
:
public void mergeSort(ArrayList<Comparable> a, int first, int last) {
int mid = (first + last)/2;
if(first == last){
}else if(last - first == 1){
// you only merge if last - first == 1...
merge(a,first, mid ,last);
}else{
last = mid;
}
}
從這一點來說,我不知道你是如何嘗試實現Merge Sort算法的。 它既不是自上而下,也不是自下而上的實現。 你在merge方法中分裂,這也很奇怪。 如果您提供了偽代碼+調用public
方法的方式,那么幫助您會更容易。 恕我直言,你的算法有一個真正的問題。
實際上,合並排序算法實現起來非常簡單。 為了說明這一點,我使用Deque
而不是List
對象編寫了這種自頂向下的合並排序算法實現 :
import java.util.Deque;
import java.util.LinkedList;
public class Example {
private LinkedList<Comparable> merge(final Deque<Comparable> left, final Deque<Comparable> right) {
final LinkedList<Comparable> merged = new LinkedList<>();
while (!left.isEmpty() && !right.isEmpty()) {
if (left.peek().compareTo(right.peek()) <= 0) {
merged.add(left.pop());
} else {
merged.add(right.pop());
}
}
merged.addAll(left);
merged.addAll(right);
return merged;
}
public void mergeSort(final LinkedList<Comparable> input) {
if (input.size() != 1) {
final LinkedList<Comparable> left = new LinkedList<Comparable>();
final LinkedList<Comparable> right = new LinkedList<Comparable>();
// boolean used to decide if we put elements
// in left or right LinkedList
boolean logicalSwitch = true;
while (!input.isEmpty()) {
if (logicalSwitch) {
left.add(input.pop());
} else {
right.add(input.pop());
}
logicalSwitch = !logicalSwitch;
}
mergeSort(left);
mergeSort(right);
input.addAll(merge(left, right));
}
}
}
我使用了Deque
因為peek()
/ pop()
比get(0)
和remove(0)
更漂亮,但它取決於你。 如果你絕對想在這里使用ArrayList
,請遵循相應的實現。
import java.util.ArrayList;
import java.util.List;
public class Example {
private List<Comparable> merge(final List<Comparable> left, final List<Comparable> right) {
final List<Comparable> merged = new ArrayList<>();
while (!left.isEmpty() && !right.isEmpty()) {
if (left.get(0).compareTo(right.get(0)) <= 0) {
merged.add(left.remove(0));
} else {
merged.add(right.remove(0));
}
}
merged.addAll(left);
merged.addAll(right);
return merged;
}
public void mergeSort(final List<Comparable> input) {
if (input.size() != 1) {
final List<Comparable> left = new ArrayList<Comparable>();
final List<Comparable> right = new ArrayList<Comparable>();
boolean logicalSwitch = true;
while (!input.isEmpty()) {
if (logicalSwitch) {
left.add(input.remove(0));
} else {
right.add(input.remove(0));
}
logicalSwitch = !logicalSwitch;
}
mergeSort(left);
mergeSort(right);
input.addAll(merge(left, right));
}
}
}
這兩種實現都適用於Integer
和String
或其他Comparable
。
希望能幫助到你。
有幾個問題,但重要的是你不應該在修改列表時迭代列表,即:
for (i = 0; i < a.size() - mid; i++){
left.add(i,a.get(i));
a.remove(i);
}
因為一旦刪除了一個元素,其他元素的索引就不一樣了......所以你添加了a
不是你想象的left
元素。
工作代碼如下(帶一些注釋):
private static void merge(ArrayList<Comparable> a) {
if (a.size()<=1) return; // small list don't need to be merged
// SEPARATE
int mid = a.size()/2; // estimate half the size
ArrayList<Comparable> left = new ArrayList<Comparable>();
ArrayList<Comparable> right = new ArrayList<Comparable>();
for(int i = 0; i < mid; i++) left.add(a.remove(0)); // put first half part in left
while (a.size()!=0) right.add(a.remove(0)); // put the remainings in right
// Here a is now empty
// MERGE PARTS INDEPENDANTLY
merge(left); // merge the left part
merge(right); // merge the right part
// MERGE PARTS
// while there is something in the two lists
while (left.size()!=0 && right.size()!=0) {
// compare both heads, add the lesser into the result and remove it from its list
if (left.get(0).compareTo(right.get(0))<0) a.add(left.remove(0));
else a.add(right.remove(0));
}
// fill the result with what remains in left OR right (both can't contains elements)
while(left.size()!=0) a.add(left.remove(0));
while(right.size()!=0) a.add(right.remove(0));
}
它已經在一些輸入上進行了測試......例如:
[4, 7, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11]
[0, 1, 1, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
為了提高效率,您可以使用subList
方法來避免顯式構造過多的子列表,它需要注意索引。
一個關於Kraal實施的警告得到了勾選標記。 這是一個很棒的實現,但是Kraal的Merge排序不保留具有相同值的項的相對順序,在某些情況下,例如在排序對象時,合並排序具有其他排序算法的重要優勢,例如quicksort , 沒有。 我修改了Kraal的代碼以保留相對訂單。
private static List<Object> merge(final List<Object> left, final List<Object> right) {
printArr("left", left);
printArr("Right", right);
final List<Object> merged = new ArrayList<>();
while (!left.isEmpty() && !right.isEmpty()) {
if(left.get(0).getValue()-right.get(0).getValue() <= 0){
merged.add(left.remove(0));
} else {
merged.add(right.remove(0));
}
}
merged.addAll(left);
merged.addAll(right);
return merged;
}
public static void mergeSort(final List<Object> input) {
if (input.size() > 1) {
final List<Object> left = new ArrayList<Object>();
final List<Object> right = new ArrayList<Object>();
boolean logicalSwitch = true;
while (!input.isEmpty()) {
if (logicalSwitch) {
left.add(input.remove(0));
} else {
right.add(input.remove(input.size()/2));
}
logicalSwitch = !logicalSwitch;
}
mergeSort(left);
mergeSort(right);
input.addAll(merge(left, right));
}
}
public class MergeSort{
public void sort(List<Integer> list){
sortAndMerge(list, 0, list.size()-1);
}
public void sortAndMerge(List<Integer> list, int start, int end){
if((end - start) >= 2){
int mid = (end - start)/2;
sortAndMerge(list, start, start + mid);
sortAndMerge(list, start + mid +1, end);
int i=start;
int j=start + mid +1;
while(i<j && j<=end){
if(list.get(i) > list.get(j)){
list.add(i, list.remove(j));
i++;
j++;
}else if(list.get(i) == list.get(j)){
list.add(i+1, list.remove(j));
i++;
j++;
}else{
i++;
}
}
}else{
if(end > start){
if(list.get(start) > list.get(end)){
int endValue = list.remove(end);
list.add(start, endValue);
}
}
}
}
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