找到两个数组之间的非常见元素

In an interview it was asked to find non- common elements between two string arrays.

在一次采访中,人们被要求在两个字符串数组之间找到非常见的元素。

Eg: String a[]={"a","b","c","d"}; 
String b[]={"b","c"}; 
O/p should be a,d

I have answered to the question that in Java Set is implemented using HashTable. The code with Set is much simpler:

我回答了Java Set中使用HashTable实现的问题。使用Set的代码更简单:

String[] a = {"a","b","c","d"};
String[] b = {"b", "c"};

Set<String> set = new HashSet<>(a.length);
for(String s : a){
    set.add(s);
}
for(String s : b){
    set.remove(s);
}
return set;

now my query is that is there any other better approach to achieve this

现在我的问题是,有没有其他更好的方法来实现这一目标

6 个解决方案

#1

You can shorten the code by

您可以缩短代码

TreeSet set = new TreeSet(Arrays.asList(a));
set.removeAll(Arrays.asList(b));

Demo

#2

If [x,y], [x,z] should yield [y,z] here's what I suggest:

如果[x,y],[x,z]应该产生[y,z],这就是我的建议:

String[] a = {"a","b","c","d"};
String[] b = {"b", "c", "x"};

Set<String> set = new HashSet<>(Arrays.asList(a));
for (String s : new HashSet<>(Arrays.asList(b)) {
    if (!set.add(s))    // if it's already present
        set.remove(s);  // remove it from the result
}

If on the other hand, [x,y], [x,z] should yield [y], I would suggest

另一方面,如果[x,y],[x,z]应该产生[y],我建议

Set<String> set = new HashSet<>(Arrays.asList(a));
set.removeAll(Arrays.asList(b));

#3

In effect, this expands upon Jon Skeet's answer, but does so using Java 8's streams.

实际上,这扩展了Jon Skeet的答案,但使用Java 8的流程。

String[] result = Arrays.stream(a)
                        .filter((s) -> Arrays.stream(b).noneMatch(s::equals))
                        .toArray(String[]::new);

System.out.println(Arrays.toString(result));

The main tenants of the code are:

该代码的主要租户是:

Filter out any element contained in A if and only if it does not exist in B (through the short-circuiting terminal operator noneMatch), checking if the element is equal to anything in that stream.

过滤掉A中包含的任何元素,当且仅当它在B中不存在时(通过短路终端运算符noneMatch),检查该元素是否等于该流中的任何元素。

Collect the results to a String[].

将结果收集到String []。

Another approach using Set, and again using streams:

使用Set的另一种方法,再次使用流:

Set<String> setA = new HashSet<>(Arrays.asList(a));
Set<String> setB = new HashSet<>(Arrays.asList(b));

String[] setResult = setA.stream()
                         .filter((s) -> !setB.contains(s))
                         .toArray(String[]::new);

The main issue with the non-Set code as pointed out was that it is quadratic time in the worst case. This code here takes advantage of the constant access time to Set#contains, and should run in about linear time.

指出非Set代码的主要问题是在最坏的情况下它是二次时间。这里的代码利用了Set#contains的持续访问时间,并且应该在大约线性时间内运行。

#4

I would handle this in three steps:

我会分三步处理:

Find all elements in a but not b

查找a中的所有元素,但不是b

Find all elements in b but not a

找到b中的所有元素但不是a

Add those two sets together

将这两个集合添加到一起

So for example:

例如:

Set<String> aSet = new HashSet<>(Arrays.asList(a));
Set<String> bSet = new HashSet<>(Arrays.asList(b));

Set<String> aNotB = new HashSet<>(aSet);
aNotB.removeAll(bSet);

Set<String> bNotA = new HashSet<>(bSet);
bNotA.removeAll(aSet);

Set<String> onlyOne = new HashSet<>(aNotB);
onlyOne.addAll(bNotA);

(The stream code in Java 8 may well make this simpler too...)

(Java 8中的流代码也可以使这更简单......)

The code could be made shorter if you don't mind modifying aSet and bSet, but I find this version easier to read.

如果您不介意修改aSet和bSet,可以缩短代码,但我发现这个版本更容易阅读。

#5

Try this:

String a[]={"a","b","c","d"}; 
String b[]={"b","c"}; 

List aLst = new ArrayList(Arrays.asList(a));
List bLst = new ArrayList(Arrays.asList(b));

aLst.removeAll(bLst);
System.out.println(aLst);

#6

If the strings are only English letters (or over a small alphabet.. even ASCII) I would rather use a boolean[] by char value instead of HashSets etc to somewhat improve performance.

如果字符串只是英文字母(或小字母......甚至是ASCII),我宁愿使用boolean [] by char值而不是HashSets等来稍微提高性能。

#1