Java 两个有序集合 求交集 差集

条件： sourceList targetList
方法一：
用java集合：
[align=left] List<TaskEntity> cloneSourceList = new ArrayList<TaskEntity>(sourceList);[/align]

[align=left] sourceList.retainAll(targetList); // 此时sourceList已是两者交集[/align]

[align=left] cloneSourceList.removeAll(sourceList); // 此时sourceList已是source差集[/align]

[align=left] targetList.removeAll(sourceList); // 此时targetList已是 target差集[/align]
[align=left]方法二：没有使用两个有序list的优越性[/align]

int sourceSize
= sourceList.size();
int targetSize
= targetList.size();
[align=left] List<T> originUpdateList = new ArrayList<T>();[/align]
for (int i
= 0; i < sourceSize; i++) {
[align=left] T source = sourceList.get(i);[/align]
boolean flag
= true ;
[align=left] T target = null;[/align]
for (int j
= 0; j < targetSize; j++) {
[align=left] target = targetList.get(j);[/align]
if (source.getID()
== target.getID()) {
[align=left] flag = false;[/align]
[align=left] originUpdateList.add( source);[/align]
if (!source.toString().equals(target.toString()))
{
[align=left] updateList.add( source);[/align]
[align=left] }[/align]
[align=left] break;[/align]
[align=left] }[/align]
[align=left] }[/align]
if (flag)
{
[align=left] insertList.add( source);[/align]
[align=left] }[/align]
[align=left] }[/align]
targetList.removeAll(originUpdateList); //
此时targetList为要删除的list
for (T
t : targetList) {
[align=left] deleteList.add(t.getID());[/align]
[align=left] }[/align]
[align=left] 此方法可以直接求出交集、sourceList的差集；但是targetList的差集还需要通过集合[/align]

[align=left]方法三：用归并排序法：[/align]

int sourceSize = sourceList.size();

int targetSize = targetList.size();

int start = 0;

for (int i = 0; i < sourceSize;) {

T source = sourceList.get(i);

for (int j = start; j < targetSize; j++) {

T target = targetList.get(j);

if (i >= sourceSize) {

deleteList.add(target.getID());

System.out.println("%^%^%" + target.getID());

continue;

}

if (source.getID() < target.getID()) {

System.out.println("insert" + source.getID() + "i" + i);

insertList.add(source);

i++;

if (i < sourceSize) {

System.out.println("oooooooooooooooooooo");

break;

} else {

deleteList.add(target.getID());

System.out.println("999999999999999999");

continue;

}

} else if (source.getID() == target.getID()) {

if (!source.toString().equals(target.toString())) {

updateList.add(source);

}

System.out.println("*********@@@@@@@@@@@@@@@@@@@@@@@@@@*" + source.getID());

i++;

if(i>=sourceSize){

continue;

}

start++;

break;

} else if (source.getID() > target.getID()) {

System.out.println("%^%^!!!!!!!!!!!!!!!!!!%" + target.getID());

deleteList.add(target.getID());

start++;

if (start >= targetSize) {

break;

}

}

}

if (start >= targetSize) {

insertList.add(source);

i++;

}

}

方法三 时间复杂度nlogn

方法二 类似n方。。。其实我也不确定