Question

此方法应该传递一个对象数组：

Movie[] movieList = new Movie[6];
    movieList[0] = new Drama("Titanic","James Cameron", 1997, 200.0, 80.0, 7.50);
    movieList[1] = new Drama("Fight Club", "David Fincher", 1999, 63.0, 30.0, 6.50);
    movieList[2] = new Animated("Spirited Away", "Hayao Miyazaki", 2001, 19.1, 2.0, 30.0);
    movieList[3] = new Animated("Toy Story", "John Lassater", 1995, 30.0, 3.5, 200.0);
    movieList[4] = new Documentary("Super Size Me","Morgan Spurlock", 2004, 0.006, 35, .005);
    movieList[5] = new Documentary("Jiro Dreams", "David Gelb", 2011, 0.003, 26, .002);

并且应该按照电影的标题进行组织和搜索。但是，每次我尝试使用switch语句将对象传递给方法时：

case 3:
    System.out.println("Please input the movie you are searching for:");
    key = input.nextLine();
    key = input.nextLine();
    if(searchMovies(movieList, key)== -1)
    {
        System.out.println("There is no match found for movie with title " + key);
    } 
    else 
    {
        index = (searchMovies(movieList, key));
        System.out.println(movieList[index].toString());
        System.out.println("\n");
    }
    break;

所有返回的都是负1，告诉我找不到密钥，或者表示数组索引超出范围的错误。这是包含冒泡排序和二进制搜索方法的searchMovies方法

/*-------------------------------------------------------------------------
//searchMovies first sorts the array of objects by title through Bubble
//Sort and then searches the array using Binary Search for the users
//key.
-------------------------------------------------------------------------*/
public static int searchMovies(Movie[] movieList, String key)
{
    //Bubble Sort the titles
    boolean needNextPass = true;
    Movie temp;
    for(int pass=1; pass<movieList.length && needNextPass; pass++)
    {
        needNextPass = false;  // Array may be sorted and next pass not needed
        for(int x=0; x<movieList.length-pass; x++)
            if(((Profitable) movieList[x]).calcProfit() < ((Profitable) movieList[x+1]).calcProfit())  /** compare rental fee */
            {
                temp = movieList[x];
                movieList[x] = movieList[x+1];
                movieList[x+1] = temp;

                needNextPass = true; // Next pass still needed
            }
     }//end for
    //Binary search for key
    int first = 0;
    int last = movieList.length;

    while (first <= last) {
        int mid =(first + last) / 2; // Compute mid point.
        if (key.compareTo(movieList[mid].getTitle()) < 0) {
            last = mid; // repeat search in bottom half.
        } else if (key.compareTo(movieList[mid].getTitle()) > 0) {
            first = mid + 1; // Repeat search in top half.
        } else {
            return mid; // Found it. return position
        }//end if
    }//end loop
    return -1; // Failed to find key
}//end searchMovies'

Answer 1

根据calcProfit()对您的电影进行冒泡，但您尝试根据getTitle()进行搜索。

如果您正在搜索的列表被视为从您用于搜索列表的比较函数的角度进行排序，则保证二进制搜索有效。在您的情况下，列表必须根据getTitle()进行排序，以便您能够使用二进制搜索。

此外，您可能获得ArrayIndexOutOfBoundsError的原因是因为您将二进制搜索的最大索引设置为moviesList.length。您应该将其设置为moviesList.length-1

Answer 2

您的电影数组需要根据title进行排序，因为您正在使用title来决定选择哪一半，以搜索下一次迭代。

Answer 3

正如其他人建议的那样，在使用二进制搜索时，您需要按照用于比较的键对数组进行排序（在您的情况下，这将是title。）

在二进制搜索算法中，您应该修改索引的分配方式如下：

int first = 0;
int last = movieList.length - 1; // last index is (length-1) in an array, not length. Trying to access length will produce an `AraryIndexOutOfBounds`

while (first <= last) {
    int mid = (first + last) / 2; 
    if (key.compareTo(movieList[mid].getTitle()) < 0) {
        last = mid - 1;  // Note the -1 here
    } else if (key.compareTo(movieList[mid].getTitle()) > 0) {
        first = mid + 1; 
    } else {
        return mid; 
    }
}

就个人而言，只有当我处于算法的学习阶段并试图弄清楚它们是如何工作的时候我才会使用上述方法。在现实生活中，我会使用类似下面的东西，因为没有必要重新发明轮子：

Movie[] movies = ....;

// How our movies are compared to each other? This comparator compares using the titles
Comparator<Movie> titleComparator = new Comparator<Movie>() {
    @Override
    public int compare(Movie o1, Movie o2) {
        return o1.title.compareTo(o2.title);
    }
};

// sort movies by title
Arrays.sort(movies, titleComparator);

String titleToFind = ...
Movie movieToFind = new UnknownMovie(titleToFind);

// perform a binary search to find that movie. If found returns the index
// if not found, returns a negative number that you can use to figure out where this movie should be inserted (see the documentation)
int index = Arrays.binarySearch(movies, movieToFind, titleComparator);

希望有所帮助

二进制搜索一直返回-1？

3 个答案: