题目：
已知一个有序数组arr 给定需要搜索的元素，返回索引，若无则返回-1

思路：
递归和非递归

1.代码如下 BinSearch .java：

package com.yuhl.right;

/**
 * @author yuhl
 * @Date 2020/10/24 21:59
 * @Classname BinSearch
 * @Description 二分查找，前提是数组为有序的，找到小搜索的元素的下标，如果不存在则返回-1
 */
public class BinSearch {
    public static void main(String[] args) {
        int[] arr = {2, 4, 6, 8};
        //非递归查询
        int index = binSearch(arr, 8);
        System.out.println(index);
        System.out.println("======================================");
        //递归查询
        int reIndex = reBinSearch(arr, 8,1,arr.length-1);
        System.out.println(reIndex);

    }

    /**
     * 采用分支的思想，O(logN)
     *
     * @param arr 数组
     * @param key 搜索值
     * @return
     */
    public static int binSearch(int[] arr, int key) {
        int low = 0;
        int high = arr.length - 1;
        int middle = 0;

        if (key < arr[low] || key > arr[high] || low > high) {
            return -1;
        }

        while (low <= high) {
            middle = (low + high) / 2;
            if (arr[middle] > key) {
                high = middle - 1;
            } else if (arr[middle] < key) {
                low = middle + 1;
            } else {
                return middle;
            }
        }
        return -1;
    }

    /**
     * 地柜二分查找
     *
     * @param arr 数组
     * @param key 搜索的值
     * @param low
     * @param high
     * @return
     */
    public static int reBinSearch(int[] arr, int key, int low, int high) {

        if (key < arr[low] || key > arr[high] || low > high) {
            return -1;
        }

        int middle = (low + high) / 2;            //初始中间位置
        if (arr[middle] > key) {
            //比关键字大则关键字在左区域
            return reBinSearch(arr, key, low, middle - 1);
        } else if (arr[middle] < key) {
            //比关键字小则关键字在右区域
            return reBinSearch(arr, key, middle + 1, high);
        } else {
            return middle;
        }
    }
}

2.执行结果：

"C:\Program Files\Java\jdk1.8.0_201\bin\java.exe" 
3
======================================
3