Arrays Equals Ignoring Order [duplicate]

问题

Possible Duplicate:
Java: Checking equality of arrays (order doesnt matter)

I have two arrays :

String[] a1 = {"a", "b", "c"};
String[] a2 = {"c", "b", "a"};

I need to check if both contains same elements (and of same length) irrespective of order of elements.

I tried Arrays.equals(a1, a2) but it considers order of element. org.apache.commons.lang.ArrayUtils does not provide this thing.

I know I can achieve the same by creating my own method (checking for same length, then sorting both array and then using Arrays.equals(a1, a2)) but wanted to know if this thing is provided in any API or there is more smart way to do the same.

回答1:

If you have these arrays in something inheriting from Collection, you can just use collection.containsAll( otherCollection ) from the Collection interface. However, you'll also need to compare the lengths of the two to verify that one set isn't a superset of the other.

(Thanks go to Aardvarkk and piegames.)

http://docs.oracle.com/javase/6/docs/api/java/util/Collection.html#containsAll(java.util.Collection)

Note: This will work up to a point. This is defined to check for at least one of any element existing. That is, if you have 3 a values in one collection, and 7 a values in the other, that will not necessarily cause it to call them unequal.

Examples:

[a, b, c] == [c, a, b]             // Works -- Mixed order
[a, b, c, d, d] == [a, b, d, c, d] // Works -- Mixed order with repeats
[a, b, c, d, d] == [a, b, b, c, d] // FAILS -- Different repeats
[a, b, c, d, d] != [a, b, c, d]    // Works -- Length differs with repeats
[a, b, c, d] != [a, b, c]          // Works -- Length differs
[a, b, d] != [a, b, c]             // Works -- Disjoint sets

回答2:

i think it may work for you, sort first array with

Arrays.sort(Object[]);

and after that you can compare with

Arrays.equals(Object[],Object[]);

complete code is:

String[] a1 = {"a", "b", "c"};
String[] a2 = {"c", "b", "a"};
Arrays.sort(a2);

boolean result= Arrays.equals(a1, a2);

回答3:

Convert the lists to sets before comparing them :

new HashSet( Arrays.asList( a1 )).equals( new HashSet( Arrays.asList( a2 ) ));

Alternatively, you can sort the arrays using Arrays.sort() but that might break code which depends on the order of the elements in the arrays.

回答4:

Use java.util.Set method equals. Compares two sets have the same size, and every member of the specified set is contained in other set.

来源：https://stackoverflow.com/questions/12009357/arrays-equals-ignoring-order

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java

arrays

equals