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给定一个规模为n的按照数字从小到大排序的数组，快速查询x元素在数组中的位置
示例：
数组：[1,3,6,9,14,35,67] 查找的值：9 输出：3
<>1、计算数据规模为n二分查找的时间复杂度
循环次数 剩下的数据规模
第一次查找： T(1) = n/2
第二次查找： T(2) = n/2^2
第三次查找： T(3) = n/2^3
…
第M次查找： T(M) = n/2^M
数据规模大于等于1即 n/2^M >=1 ，说明不能再继续二分查找的操作，当数据规模达到最小值1时即n/2^M =1则是最坏的查找情况。
T(M) = n/2^M = 1 得到M=O(N)=log2n，二分查找的时间复杂度为以2为底数n为指数的对数。
<>2、算法如下
package com.example.algorithm; /** * 给定一个规模为n的按照数字从小到大排序的数组，快速查询x元素在数组中的位置 *
<p> * 示例： * <p> * <p> * 数组：[1,3,6,9,14,35,67] * * <p> * <p> * 查找的值：9 * <p> *
<p> * 输出：3 * * @author xukaijun5 * @since 2021/11/2 **/ public class Day20211102
{ public static int getIndex(int[] source, int target) { int left = 0; int right
= source.length - 1; int mid; while (left <= right) { mid = (left + right) >> 1;
if (source[mid] == target) { return mid; } else if (source[mid] > target) {
right= mid - 1; } else { left = mid + 1; } } return -1; } public static void
main(String[] args) { int[] source = new int[]{1, 3, 6, 9, 14, 35, 67}; System.
out.println(getIndex(source, 9)); } }
