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How to check if an element is in an array?

How to check if an element is in an array?# Java - 爪哇娇娃

y*o2006-11-23 08:11

1 楼

In ArrayList, we can use contains(Object o) which returns a boolean value.
How about an array of primitive integers? In Python, we can do
if x in myList:
print "I got you!"
Any easy way to do the same in Java? I have always been wondering about
this.
Thanks.

B*h2006-11-23 08:11

2 楼

==

【在 y********o 的大作中提到】

: In ArrayList, we can use contains(Object o) which returns a boolean value.
: How about an array of primitive integers? In Python, we can do
: if x in myList:
: print "I got you!"
: Any easy way to do the same in Java? I have always been wondering about
: this.
: Thanks.

g*g2006-11-23 08:11

3 楼

People usually do a for loop to check,
boolean found = false;
for(i=0; iif(o.equals(array[i]) {
found = true;
break;
}
}
if you are really lazy, try Arrays.asList(array).contains()
For trivial array, shouldn't be much different in performance.

【在 y********o 的大作中提到】

y*o2006-11-23 08:11

4 楼

Thanks a lot! Yes, I also do for loop, but I had thought that there must be
some smarter way of doing that. The one you offered looks good.
发信人: goodbug (好虫), 信区: Java
标题: Re: How to check if an element is in an array?
发信站: BBS 未名空间站 (Thu Nov 23 11:38:48 2006), 转信
People usually do a for loop to check, if you are really lazy,
try Arrays.asList(array).contains()

m*t2006-11-23 08:11

5 楼

java.util.Arrays.binarySearch(int[], int)

【在 y********o 的大作中提到】

g*g2006-11-23 08:11

6 楼

binary search will require a sorting in advance, which probably is not
what you want.

【在 m******t 的大作中提到】

: java.util.Arrays.binarySearch(int[], int)

o*t2006-11-23 08:11

7 楼

even the for loop has more lines, the performance is the same. ArrayList.
contains just do the same thing. In Java 5, we can use foreach

【在 y********o 的大作中提到】

y*o2006-11-23 08:11

8 楼

其实我在考虑这个问题是因为我想生成比方说100个random integer, 要个个不同。所以
想，生成一个放入array, 往后生成的，先看看array里面是否已经存在，是新鲜的才放
进去。
下午又想起来，可能可以用HashSet，将生成的random integer拿Integer包了放入Hash
Set里面去，这样就省了比较这一步了。
如果java的Collection classes可以接收primitive type就好了。像上面用HashSet的办
法，到头来还要打开包转换一下方能使用，不是很麻烦吗？
当然，如果根本就有极其简单的办法得到100个unique random integers, 请赐教。我不
熟java。

【在 o******t 的大作中提到】

: even the for loop has more lines, the performance is the same. ArrayList.
: contains just do the same thing. In Java 5, we can use foreach

y*o2006-11-23 08:11

9 楼

I just tried the new "for" syntax for a hashset like this:
for (String mystrElement: myHashSet)
It worked. Hmm, much better than getting the iterator and then do hasNext().
Is this "for" syntax supposed to work for all Collection classes?

【在 o******t 的大作中提到】

: even the for loop has more lines, the performance is the same. ArrayList.
: contains just do the same thing. In Java 5, we can use foreach

B*h2006-11-23 08:11

10 楼

arrays too.

().

【在 y********o 的大作中提到】

: I just tried the new "for" syntax for a hashset like this:
: for (String mystrElement: myHashSet)
: It worked. Hmm, much better than getting the iterator and then do hasNext().
: Is this "for" syntax supposed to work for all Collection classes?

r*l2006-11-23 08:11

11 楼

You should definitely use Set instead of Array to do the job. Java 1.5 has
auto boxing and un-boxing so you don't have to do the conversion.

所以
Hash
的办
我不

【在 y********o 的大作中提到】

: 其实我在考虑这个问题是因为我想生成比方说100个random integer, 要个个不同。所以
: 想，生成一个放入array, 往后生成的，先看看array里面是否已经存在，是新鲜的才放
: 进去。
: 下午又想起来，可能可以用HashSet，将生成的random integer拿Integer包了放入Hash
: Set里面去，这样就省了比较这一步了。
: 如果java的Collection classes可以接收primitive type就好了。像上面用HashSet的办
: 法，到头来还要打开包转换一下方能使用，不是很麻烦吗？
: 当然，如果根本就有极其简单的办法得到100个unique random integers, 请赐教。我不
: 熟java。

c*t2006-11-23 08:11

12 楼

autoboxing should be avoided.

【在 r*****l 的大作中提到】

: You should definitely use Set instead of Array to do the job. Java 1.5 has
: auto boxing and un-boxing so you don't have to do the conversion.
:
: 所以
: Hash
: 的办
: 我不

m*t2006-11-23 08:11

13 楼

Right. For some reason I took for granted the OP was talking about a sorted
array.

【在 g*****g 的大作中提到】

: binary search will require a sorting in advance, which probably is not
: what you want.