如何从两个数组列表中删除公共值
如何从两个ArrayList中删除公共值。 让我们考虑我有两个数组列表如下图所示如何从两个数组列表中删除公共值
ArrayList1= [1,2,3,4]
ArrayList1= [2,3,4,6,7]
我想有结果作为
ArrayListFinal= [1,6,7]
任何人都可以请帮我吗?
这里是你可以按照完成任务的算法:
- 构建两个数组的联合
- 构建两个数组的交集
- 从工会减去路口让你的结果
Java集合支持addAll
,removeAll
和retainAll
。使用addAll
建设工会,retainAll
构建交叉口,和removeAll
作减法,like this:
// Make the two lists
List<Integer> list1 = Arrays.asList(1, 2, 3, 4);
List<Integer> list2 = Arrays.asList(2, 3, 4, 6, 7);
// Prepare a union
List<Integer> union = new ArrayList<Integer>(list1);
union.addAll(list2);
// Prepare an intersection
List<Integer> intersection = new ArrayList<Integer>(list1);
intersection.retainAll(list2);
// Subtract the intersection from the union
union.removeAll(intersection);
// Print the result
for (Integer n : union) {
System.out.println(n);
}
SetList<Integer> A = new SetList<Integer>();
A.addAll({1,2,3,4});
SetList<Integer> B = new SetList<Integer>();
B.addAll({2,3,4,6,7});
Integer a = null;
for (int i=0; i<A.size(); i++)
{
a = A.get(i);
if (B.contains(a)
{
B.remove(a);
A.remove(a);
i--;
}
}
SetList<Integer> final = new SetList<Integer>();
final.addAll(A);
final.addAll(B);
// final = { 1, 6, 7 }
你实际上问了Symmetric Difference。
List<Integer> aList = new ArrayList<>(Arrays.asList(1, 2, 3, 4));
List<Integer> bList = new ArrayList<>(Arrays.asList(2, 3, 4, 6, 7));
// Union is all from both lists.
List<Integer> union = new ArrayList(aList);
union.addAll(bList);
// Intersection is only those in both.
List<Integer> intersection = new ArrayList(aList);
intersection.retainAll(bList);
// Symmetric difference is all except those in both.
List<Integer> symmetricDifference = new ArrayList(union);
symmetricDifference.removeAll(intersection);
System.out.println("aList: " + aList);
System.out.println("bList: " + bList);
System.out.println("union: " + union);
System.out.println("intersection: " + intersection);
System.out.println("**symmetricDifference: " + symmetricDifference+"**");
打印:
aList: [1, 2, 3, 4]
bList: [2, 3, 4, 6, 7]
union: [1, 2, 3, 4, 2, 3, 4, 6, 7]
intersection: [2, 3, 4]
**symmetricDifference: [1, 6, 7]**
感谢您的回复...这就是我一直在寻找:) :) – Gautam 2013-03-22 18:58:59
@OldCurmudgeon如果我想要一个像[1,2,3,4,6,7]这样的输出,我该怎么办? – 2015-02-02 12:42:13
@ KK_07k11A0585 - 这是“对称差异”和“交集”的“联合”。 – OldCurmudgeon 2015-02-02 12:45:38
您可以使用这样的事情:
ArrayList <Integer> first = new ArrayList <Integer>();
ArrayList <Integer> second = new ArrayList <Integer>();
ArrayList <Integer> finalResult = new ArrayList <Integer>();
first.add(1);
first.add(2);
first.add(3);
first.add(4);
second.add(2);
second.add(3);
second.add(4);
second.add(6);
second.add(7);
for (int i = 0; i < first.size(); i++){
if (!second.contains(first.get(i))){
finalResult.add(first.get(i));
}
}
for (int j = 0; j < second.size(); j++){
if (!first.contains(second.get(j))){
finalResult.add(second.get(j));
}
}
我只是填充2周的ArrayList如您在您的文章中描述他们,我对他们两个检查针对不同的元素;如果找到了这样的元素,我将它们添加到finalResult ArrayList中。
我希望它会帮助你:)
试试看,那么我们会帮助你。 – 2013-03-22 16:34:27
所以你想删除常见元素 – SRy 2013-03-22 16:35:49
提示:如果你阅读列表的API,你会自己解决你的问题。 – Sanchit 2013-03-22 16:36:41