[英]merge sort list java
此代碼的輸出始終是輸入的最后一位。 找不到原因。我遞歸地使用合並排序,結果是錯誤的。 我以為清單可能重疊了。
public class MergeSort {
public static List<Integer> Sort(List<Integer> list) {
if (list.size() <= 1) {
return list;
}
List<Integer> aList = new ArrayList<Integer>();
aList = list.subList(0, list.size() / 2);
List<Integer> bList = new ArrayList<Integer>();
bList = list.subList(list.size() / 2, list.size());
Sort(aList);
Sort(bList);
merge(aList, bList, list);
return list;
}
private static List<Integer> merge(List<Integer> alist,
List<Integer> blist, List<Integer> list) {
int alistIndex = 0, blistIndex = 0, listIndex = 0;
while (alistIndex < alist.size() && blistIndex < blist.size()) {
if (alist.get(alistIndex) < blist.get(blistIndex)) {
list.set(listIndex, alist.get(alistIndex));
alistIndex++;
} else {
list.set(listIndex, blist.get(blistIndex));
blistIndex++;
}
listIndex++;
}
List<Integer> rest;
if (alistIndex == alist.size()) {
rest = blist.subList(blistIndex, blist.size());
for(int c = blistIndex; c < rest.size(); c++){
list.set(listIndex, blist.get(c));
listIndex++;
}
} else {
rest = alist.subList(alistIndex, alist.size());
for(int c = alistIndex; c < rest.size(); c++){
list.set(listIndex, alist.get(c));
listIndex++;
}
}
return list;
}
}
測試輸入為5、4、3、2、1。但是輸出為1、1、1、1、1。因此,此合並方法一定存在一些錯誤。
解決您的問題的快速方法是更換:
List<Integer> aList = new ArrayList<Integer>();
aList = list.subList(0, list.size() / 2);
List<Integer> bList = new ArrayList<Integer>();
bList = list.subList(list.size() / 2, list.size());
有:
List<Integer> aList = new ArrayList<Integer>(list.subList(0, list.size() / 2));
List<Integer> bList = new ArrayList<Integer>(list.subList(list.size() / 2, list.size()));
您為分區創建新的ArrayList
,然后立即將引用更改為原始列表的視圖。
列表的分區正確完成了,但是,由於您使用的是視圖而不是淺表副本,因此在合並過程中,您正在更改分區。
通常,如果您要進行的操作使原始列表發生突變,則不會從該方法返回任何內容,因此類似:
public class MergeSort {
public static void sort(List<Integer> list) {
if (list.size() < 2) {
return;
}
int mid = list.size()/2;
List<Integer> left = new ArrayList<Integer>(list.subList(0, mid));
List<Integer> right = new ArrayList<Integer>(mid, list.size()));
sort(left);
sort(right);
merge(left, right, list);
}
private static void merge(
List<Integer> left, List<Integer> right, List<Integer> list) {
int leftIndex = 0;
int rightIndex = 0;
int listIndex = 0;
while (leftIndex < left.size() && rightIndex < right.size()) {
if (left.get(leftIndex) < right.get(rightIndex)) {
list.set(listIndex++, left.get(leftIndex++));
} else {
list.set(listIndex++, right.get(rightIndex++));
}
}
while (leftIndex < left.size()) {
list.set(listIndex++, left.get(leftIndex++));
}
while (rightIndex < right.size()) {
list.set(listIndex++, right.get(rightIndex++));
}
}
}
不更改原始列表的替代方法可能是:
public class MergeSort {
public static List<Integer> sorted(List<Integer> list) {
if (list.size() < 2) {
return list;
}
int mid = list.size()/2;
return merged(
sorted(list.subList(0, mid)),
sorted(list.subList(mid, list.size())));
}
private static List<Integer> merged(List<Integer> left, List<Integer> right) {
int leftIndex = 0;
int rightIndex = 0;
List<Integer> merged = new ArrayList<Integer>();
while (leftIndex < left.size() && rightIndex < right.size()) {
if (left.get(leftIndex) < right.get(rightIndex)) {
merged.add(left.get(leftIndex++));
} else {
merged.add(right.get(rightIndex++));
}
}
merged.addAll(left.subList(leftIndex, left.size()));
merged.addAll(right.subList(rightIndex, right.size()));
return merged;
}
}
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