java 常用算法

选择排序，复杂度O（n²）

package com.example.demo;

import org.junit.Test;
/**
 * 选择排序
 * @author zhzh.yin
 *
 */
public class HTest {
	@Test
	public void testMethod(){
		Integer [] numList = {1,2,2,2,1,4,5,2,5,3,9,6};
		for(int i=0;i<numList.length-1;i++){
			int max = numList[i];
			for(int j = i+1;j<numList.length;j++){
				if(numList[i]<numList[j]){
					max = numList[j];
					numList[j]=numList[i];
					numList[i]=max;
				}
			}
		}
		for(int num:numList){
			System.out.print(numList[num]+" ");
		}
	}
}

冒泡排序复杂度 O(nlogn)

package com.example.demo;

import org.junit.Test;
/**
 * 冒泡排序
 * @author zhzh.yin
 *
 */
public class FTest {
	@Test
	public void testMethod(){
		Integer[]numList = {1,2,1,1,1,8,3,5,1};
		if(numList.length==1){
			System.out.println("-1");
		}else if(numList.length==2){
			String string ="";
			System.out.println(
					numList[1]>numList[2]?
					numList[1]+""+numList[0]:numList.toString());;
		}
		int temp = 0;
		for(int i =0;i<numList.length-1;i++){
			for(int j=i+1;j<numList.length;j++){
				if(numList[i]<numList[j]){
					temp=numList[i];
					numList[i]=numList[j];
					numList[j]=temp;
				}
			}
		}
		for(int num :numList){
			System.out.print(num+" ");
		}
	}
}

二分法查询-非迭代

import org.junit.Test;
/**
 * 找到数字N，适用于有多个重复数字，使用部分math原生方法向上取整/向下取整
 * middle=min+(max-min)/2 算法更加准确
 * 返回-1为没有找到
 * @author zhzh.yin
 *
 */
public class CTest {
	@Test
	public void testMethod(){
		Integer[] num ={1,2,3,3,3,4,5,5,5,5,5,6,6,6,7,10};
		System.out.println("the num is at "+selectPosition(num,8));
	}
	public int selectPosition(Integer [] numList,int num){
		int min=0;
		int max=numList.length-1;
		int middle = -1;
		while(min<=max){
			middle = min+(max-min)/2;
			System.out.println("min--"+min+",max--"+max+",middle--"+middle);
			if((min==middle||max==middle)&&numList[middle]!=num){
				middle = -1;
				break;
			}
			if(numList[middle]<num){
				min=(int) Math.floor(middle);
			}else if(numList[middle]>num){
				max=(int) Math.ceil(middle);
			}else{
				break;
			}
		}
		return middle;
	}
}

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