java通过二分法对数组进行高效查找及插入操作

最新推荐文章于 2022-11-16 21:55:19 发布

原创最新推荐文章于 2022-11-16 21:55:19 发布 · 1.2k 阅读

2 ·

CC 4.0 BY-SA版权

文章标签：

#java #class

Java基础同时被 2 个专栏收录

16 篇文章

订阅专栏

Java小算法

5 篇文章

订阅专栏

本文介绍了一个有序数组的Java类实现，包括插入、查找、删除等基本操作。通过二分查找提高搜索效率，并维持数组有序性。

public class OrderArray {
    private long[] a;
    private int nElems;

    public OrderArray(int max){
        a = new long[max];
        nElems = 0;//代表数组中元素个数
    }

    public int size(){
        return nElems;
    }

    public int find(long searchKey){
        int beginIndex = 0;
        int endIndex = nElems - 1;
        int center;
        while (true){
            center = (beginIndex + endIndex)/2;
            if(a[center] == searchKey)
                return center;
            else if(beginIndex > endIndex){
                return nElems;
            }else {
                if(a[center] < searchKey)
                    beginIndex = center + 1;
                else
                    endIndex = center - 1;
            }
        }
    }

    public boolean insert(long value){
        nElems = find(value);
        if(nElems == 0 || value > a[nElems - 1]){
            a[nElems] = value;
            nElems++;
            return true;
        }
        if(value < a[0]) {
            for (int k = nElems; k > 0; k--)
                a[k] = a[k - 1];
            a[0] = value;
            nElems++;
            return true;
        }

        int beginIndex = 0;
        int endIndex = nElems - 1;
        int center;
        while (true){
            center = (beginIndex + endIndex)/2;
            if(a[center] < value){
                if(a[center + 1] > value){
                    for(int k = nElems;k > center + 1;k--)
                        a[k] = a[k - 1];
                    a[center + 1] = value;
                    nElems++;
                    return true;
                }else{
                    beginIndex = center + 1;
                }
            }else{
                    endIndex = center;
            }
        }
    }

    public boolean delete(long value){
        int i = find(value);
        if(i == nElems)
            return false;
        else {
            for (; i < nElems; i++)
                a[i] = a[i + 1];
            nElems--;
            return true;
        }
    }

    public void display(){
        for(int j = 0;j < nElems;j++)
            System.out.print(a[j] + " ");
        System.out.println(" ");
    }
}