如何使用二进制搜索处理数组中的重复数字和两位或多位数字

Question

我写了一个二叉搜索算法，它工作正常，但是当数组中有重复项或对于数字两位或更多位数字时，它无法正常工作，它返回数组中找不到的值。我能做些什么来解决这个问题，我是否需要在数组中搜索任何重复项来处理重复项？或者我可以阻止随机数生成器生成重复项吗？

这是代码：

public class BinarySearch {
    static int RandomNum;
    static int NumSize;
    static int[] arr;
    static int rangeOfNum;
    static int LookFor;
    static Random numGen = new Random();
    static Scanner Input = new Scanner(System.in);
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        System.out.println("Enter the size of the data set");
        NumSize = Input.nextInt();
        System.out.println("Enter the range of the random numbers");
        rangeOfNum = Input.nextInt();
        arr = new int[NumSize];
        for (int i = 0; i <= NumSize-1; i++) {
            RandomNum = numGen.nextInt(rangeOfNum);
            arr[i] = RandomNum; 
        }
        Arrays.sort(arr);
        for (int i = 0; i < arr.length; i++)
        System.out.println("Number is: " + arr[i]);
        System.out.println("Enter number you want to search for: ");
        LookFor = Input.nextInt();
        binarySearch(arr[0], arr.length-1, LookFor, arr);
    }
    //Binary Search Algorithm
    public static int binarySearch(int LowIndex, int HighIndex, int Target, int[] array) {  
        if (LowIndex <= HighIndex) {
            int MidIndex = (LowIndex + HighIndex)/2;
            if (array[MidIndex] == Target){
                System.out.println("Found target value at index " + MidIndex);
                return Target;}
                else if (array[MidIndex] > Target)
                    return binarySearch(LowIndex, MidIndex-1, Target, array);
                    else 
                        return binarySearch(MidIndex+1, HighIndex, Target, array);
        }
        System.out.println("target value not found");
        return -1;
    }
}

感谢您的帮助。

Answer 1

第一次调用binarySearch时，您按如下方式调用它：

binarySearch(arr[0], arr.length-1, LookFor, arr);

将其更改为

binarySearch(0, arr.length-1, LookFor, arr);