如何获取数组列表的最后一个值?


当前回答

这对我很管用。

private ArrayList<String> meals;
public String take(){
  return meals.remove(meals.size()-1);
}

其他回答

这个怎么样? 在你班上的某个地方……

List<E> list = new ArrayList<E>();
private int i = -1;
    public void addObjToList(E elt){
        i++;
        list.add(elt);
    }


    public E getObjFromList(){
        if(i == -1){ 
            //If list is empty handle the way you would like to... I am returning a null object
            return null; // or throw an exception
        }

        E object = list.get(i);
        list.remove(i); //Optional - makes list work like a stack
        i--;            //Optional - makes list work like a stack
        return object;
    }

如果可以的话,将ArrayList替换为ArrayDeque,它有方便的方法,如removeLast。

我使用micro-util类获取列表的最后(和第一个)元素:

public final class Lists {

    private Lists() {
    }

    public static <T> T getFirst(List<T> list) {
        return list != null && !list.isEmpty() ? list.get(0) : null;
    }

    public static <T> T getLast(List<T> list) {
        return list != null && !list.isEmpty() ? list.get(list.size() - 1) : null;
    }
}

稍微灵活一点:

import java.util.List;

/**
 * Convenience class that provides a clearer API for obtaining list elements.
 */
public final class Lists {

  private Lists() {
  }

  /**
   * Returns the first item in the given list, or null if not found.
   *
   * @param <T> The generic list type.
   * @param list The list that may have a first item.
   *
   * @return null if the list is null or there is no first item.
   */
  public static <T> T getFirst( final List<T> list ) {
    return getFirst( list, null );
  }

  /**
   * Returns the last item in the given list, or null if not found.
   *
   * @param <T> The generic list type.
   * @param list The list that may have a last item.
   *
   * @return null if the list is null or there is no last item.
   */
  public static <T> T getLast( final List<T> list ) {
    return getLast( list, null );
  }

  /**
   * Returns the first item in the given list, or t if not found.
   *
   * @param <T> The generic list type.
   * @param list The list that may have a first item.
   * @param t The default return value.
   *
   * @return null if the list is null or there is no first item.
   */
  public static <T> T getFirst( final List<T> list, final T t ) {
    return isEmpty( list ) ? t : list.get( 0 );
  }

  /**
   * Returns the last item in the given list, or t if not found.
   *
   * @param <T> The generic list type.
   * @param list The list that may have a last item.
   * @param t The default return value.
   *
   * @return null if the list is null or there is no last item.
   */
  public static <T> T getLast( final List<T> list, final T t ) {
    return isEmpty( list ) ? t : list.get( list.size() - 1 );
  }

  /**
   * Returns true if the given list is null or empty.
   *
   * @param <T> The generic list type.
   * @param list The list that has a last item.
   *
   * @return true The list is empty.
   */
  public static <T> boolean isEmpty( final List<T> list ) {
    return list == null || list.isEmpty();
  }
}

如解决方案中所述,如果List为空,则抛出IndexOutOfBoundsException。一个更好的解决方案是使用Optional类型:

public class ListUtils {
    public static <T> Optional<T> last(List<T> list) {
        return list.isEmpty() ? Optional.empty() : Optional.of(list.get(list.size() - 1));
    }
}

如你所料,列表的最后一个元素作为Optional返回:

var list = List.of(10, 20, 30);
assert ListUtils.last(list).orElse(-1) == 30;

它还可以优雅地处理空列表:

var emptyList = List.<Integer>of();
assert ListUtils.last(emptyList).orElse(-1) == -1;

在Kotlin中,你可以使用最后的方法:

val lastItem = list.last()