如果 x 在 Java 中的数组中
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If x in array in Java
提问by user3312175
In python, you can use a very simple if statement to see if what the user has entered (or just a variable) is in a list:
在 python 中，您可以使用一个非常简单的 if 语句来查看用户输入的内容（或只是一个变量）是否在列表中：
myList = ["x", "y", "z"]
myVar = "x"
if myVar in x:
print("your variable is in the list.")
How would I be able to do this in Java?
我如何才能在 Java 中做到这一点？
采纳答案by Sotirios Delimanolis
If your array type is a reference type, you can convert it to a Listwith Arrays.asList(T...)and check if it contains the element
如果您的数组类型是引用类型，则可以将其转换为ListwithArrays.asList(T...)并检查它是否包含该元素
if (Arrays.asList(array).contains("whatever"))
// do your thing
回答by DmitryKanunnikoff
String[] array = {"x", "y", "z"};
if (Arrays.asList(array).contains("x")) {
System.out.println("Contains!");
}
回答by Elliott Frisch
Here is one solution,
这是一种解决方案，
String[] myArray = { "x", "y", "z" };
String myVar = "x";
if (Arrays.asList(myArray).contains(myVar)) {
System.out.println("your variable is in the list.");
}
Output is,
输出是，
your variable is in the list.
回答by Brian
You could iterate through the array and search, but I recommend using the SetCollection.
您可以遍历数组并进行搜索，但我建议使用SetCollection。
Set<String> mySet = new HashSet<String>();
mySet.add("x");
mySet.add("y");
mySet.add("z");
String myVar = "x";
if (mySet.contains(myVar)) {
System.out.println("your variable is in the list");
}
Set.contains()is evaluated in O(1)where traversing an arrayto search can take O(N)in the worst case.
Set.contains()在最坏的情况下O(1)遍历arrayto search 的位置进行评估O(N)。
回答by Travis O'Donnell
Rather than calling the Arrays class, you could just iterate through the array yourself
您可以自己遍历数组，而不是调用 Arrays 类
String[] array = {"x", "y", "z"};
String myVar = "x";
for(String letter : array)
{
if(letter.equals(myVar)
{
System.out.println(myVar +" is in the list");
}
}
Remember that a String is nothing more than a character array in Java. That is
请记住，字符串只不过是 Java 中的字符数组。那是
String word = "dog";
is actually stored as
实际上存储为
char[] word = {"d", "o", "g"};
So if you would call if(letter == myVar), it would never return true, because it is just looking at the reference id inside the JVM. That is why in the code above I used
因此，如果您调用 if(letter == myVar)，它将永远不会返回 true，因为它只是查看 JVM 中的引用 ID。这就是为什么在上面的代码中我使用
if(letter.equals(myVar)) { }
