先看看我们用 contain 去重的代码: /** * 使用 list.contain 去重 * * @param testList */ private static void useContain2Distinct(List<String> testList) { System.out.println("contains 开始去重,条数:" + testList.size()); List<String> testListDistinctResult = new ArrayList<>(); for (Strin...
Java中list连接 java list contain 在实际项目中我们通常会有一个需求就是:想知道在一个列表中是否包含某一个对象 这里ArrayList表、HashSet表和HashMap表都提供了一个contains(obj)方法, 下面说一下两个列表contains(obj)方法的实现原理。 ArrayList表: 先遍历表中每个元素(对象),然后对每个元素执行一个equals(obj...
在Java中,要查看一个List是否包含某个字符串,可以使用List的contains()方法。这个方法会返回一个boolean值,表示该List是否包含指定的元素。 例如,假设有一个List list,可以通过以下代码来判断是否包含某个字符串: List<String> list = new ArrayList<>(); list.add("apple"); list.add("banana"); list.add(...
print("No, it does not contain orange") 执行以上代码输出结果为: No, it does not contain orange list.contains 方法是一种非常常用的元素查找的方式,它的优点在于可以很快的检 查列表中元素是否存在于某个 List 中。特别适用于大规模的列表查找,这样可以避免不 必要的循环操作,节省时间,提高效率。 Java中...
It's a matter of logic: does A contain all the elements inside B? This can be seen as for each element in B, does this element belong to A too? You can understand that the condition is true, since B is empty, there is no element to check: for each element in B, so for no ...
Think about the following scenario: If I have a List<String>, it is supposed to only contain objects of type String. (Which is a final class) If I can cast that to a List<Object>, then that allows me to add Object to that list, thus violating the original contract of List<Strin...
checkFruit("grape") // Output: No, the list does not contain grape. checkFruit("watermelon") // Output: The fruit name is longer than 5 characters. } 在上述示例中,我们首先创建了一个包含若干水果名称的列表fruits。然后定义了一个名为checkFruit的函数,该函数接受一个水果名称作为参数。在函数体...
* in this list, or -1 if this list does not contain the element. * More formally, returns the highest index i such that * (o==null ? get(i)==null : o.equals(get(i))), * or -1 if there is no such index.*/publicintlastIndexOf(Object o){if(o ==null) {for(inti = size...
/*** Removes the first occurrence of the specified element from this list,* if it is present. If this list does not contain the element, it is* unchanged. More formally, removes the element with the lowest index* {@code i} such that* (o==null ? get(i)==null : o.equals(get(i)...
The set is contain 3 and 4 1 2 3 4 6 12346 TreeSet 基于TreeMap,生成一个总是处于排序状态的set,内部以TreeMap来实现。它是使用元素的自然顺序对元素进行排序,或者根据创建Set 时提供的 Comparator 进行排序,具体取决于使用的构造方法。 publicstaticvoidmain(String[] args) { ...