Question

我需要在具有各种数据类型的链表上执行二进制搜索。下面的代码将无法编译。我似乎无法使compareTo（）工作。

这是链表类：

public class Contributor {

private String firstName;
private String lastName;
private String country;
private String phone;
private double contribution;
private int id;}

二进制搜索方法如下。搜索需要使用二进制搜索方法找到某个lastName。

 public void binarySearch(List<Contributor> l, String key) {
    System.out.println("Binary search.");

    int upperBound = l.size();
    int lowerBound = 1;
    int midpoint = (upperBound + lowerBound) / 2;
    int difference = upperBound - lowerBound;

    for (int i = 0; i < l.size(); i++) {
        if (key.compareTo(l.get(midpoint - 1))&& difference != 1) {
            upperBound = midpoint - 1;
            midpoint = upperBound / 2;
        } else if (key.compareTo(l.get(midpoint - 1)) && difference != 1) {
            lowerBound = midpoint + 1;
            midpoint = (lowerBound + upperBound) / 2;

        } else if (key.equals(l.get(midpoint - 1))) {
            midpoint = midpoint - 1;

            System.out.println("We found " + key + " at position " + midpoint + " in the list.");
            i = l.size();
        } else {
            System.out.println("We couldn't find " + key + " in the list.");
            i = l.size();
        }
    }
}

Answer 1

通过查看代码，l.get（midpoint-1）返回的是Contributor对象而不是String，而key.compareTo方法只能接受String类型的参数。

Answer 2

compareTo（）函数返回整数值。所以试试

if(key.compareTo(str) < 0)

或

if(key.compareTo(str) > 0)

或

if(key.compareTo(str) == 0)

还要将它与Contributor类中的String进行比较，而不是与类本身进行比较。

if(key.compareTo((l.get(midpoint - 1)).firstName)<0)

另一个逻辑错误是当你做

时

midpoint = upperBound / 2;

您假设lowerbound 1 。但它的价值可能会在迭代中发生变化。所以考虑一下

Answer 3

这里有一个逻辑错误 - 正如Akshay所说，你的密钥是String，因此在Contibutor列表中搜索它是没有意义的 - 你需要搜索对于Contributor。

除此之外，二进制搜索的前提条件是列表已排序，这要求您的Contributor对象为Comparable（或者至少提供Comparator。

我已修改您的代码以解决这些问题：

class Contributor implements Comparable<Contributor>{

    private String firstName;
    private String lastName;
    private String country;
    private String phone;
    private double contribution;
    private int id;

    public int compareTo(Contributor o) {
        return Integer.valueOf(id).compareTo(o.id);
    }
}

public class BinarySearch {
    /**
     * Performs a binary search for <code>key</code> in the list <code>l</code>.
     * 
     * @param l Ordered list of Contributors.
     * @param key Value to search for.
     */
    public void binarySearch(List<Contributor> l, Contributor key) {
        System.out.println("Binary search.");

        int upperBound = l.size();
        int lowerBound = 1;
        int midpoint = (upperBound + lowerBound) / 2;
        int difference = upperBound - lowerBound;

        for (int i = 0; i < l.size(); i++) {
            if (key.compareTo(l.get(midpoint - 1)) <0 ) {
                upperBound = midpoint - 1;
                midpoint = upperBound / 2;
            } else if (key.compareTo(l.get(midpoint - 1))>0 ) {
                lowerBound = midpoint + 1;
                midpoint = (lowerBound + upperBound) / 2;

            } else if (key.equals(l.get(midpoint - 1))) {
                midpoint = midpoint - 1;

                System.out.println("We found " + key + " at position "
                        + midpoint + " in the list.");
                i = l.size();
            } else {
                System.out.println("We couldn't find " + key + " in the list.");
                i = l.size();
            }
        }
    }
}

这应该可以修复你的编译错误，尽管其他人注意到，搜索逻辑本身也有一些错误。另请注意，binarySearch方法的前提条件是订购列表，并且应记录此要求。我添加了一个javadoc注释来解释这个。

链接列表中字符串的二进制搜索

3 个答案: