二分法查找数组中元素

要使用二分法需要注意:  数组中的元素必须已经按升序排列好

二分法主要思想是将一个数组一分为二，每次查询都能将查询范围在上一次的基础上缩小一半。所以效率非常高。

下面是Java代码实现:

import java.util.*; 

public class Main
{

    public static void main(String[] args) throws Exception
    {
        Integer[] numbers = new Integer[] 
        {
                1,3,2,1,2,3,2,7,7,8,2
        };
        Integer target = 7;
        Integer head = 0;
        Integer tail = numbers.length - 1;
        Integer middle = (head + tail)/2;
        
        Arrays.sort(numbers); 
        for(Integer item : numbers)
            System.out.print(item + " ");
        System.out.println();
        Integer targetIndex = dichotomySearch(numbers,head,middle,tail,target);
        if(targetIndex == null)
            System.out.println("target not found");
        else
            System.out.println("target in: " + targetIndex);
    }
    
    public static Integer dichotomySearch(Integer[] numbers,Integer head, Integer middle,Integer tail,Integer target)
    {
        //System.out.println("head:" + head + " middle:" + middle + " tail:" + tail);
        if(target.equals(numbers[head]))
            return head;
        if(target.equals(numbers[middle]))
            return middle;
        if( target.equals(numbers[tail]))
            return tail;
        
        
        if(head.equals(middle) || head.equals(tail)||middle.equals(tail))
            return null;
        
        if(target.compareTo(numbers[head]) > 0 && target.compareTo(numbers[middle]) < 0)
        {
            System.out.println("A");
            tail = middle;
            middle = (head + tail)/2;
        }
        else if(target.compareTo(numbers[middle]) > 0 && target.compareTo(numbers[tail]) < 0)
        { 
            head = middle;
            middle = (head + tail)/2; 
        }
        else
        {
            return null;
        }
        
        return dichotomySearch(numbers, head,  middle, tail, target);
    } 
}

输出结果:

1 1 2 2 2 2 3 3 7 7 8
target in: 8