偶然在群里看到有人问到大数据查询，自己也就想了小艾改如何解决，从从1000万个随机数中查找出相同的10万个随机数花的最少时间，
谈到效率，自然是hashmap莫属。
import java.util.HashMap;
import java.util.Map;
import java.util.Random;
public class dataTest {
private static final int [] datas = new int [1000 * 10000]; //1000万个随机数
private static final int [] params = new int [10 * 10000]; // 10万个随机数
private static final Map<Integer, Integer> map = new HashMap<Integer, Integer>();
private static final Random random = new Random(System.currentTimeMillis());
static {
int length = datas.length;
//初始化1000万个数字
for(int i = 0; i < datas.length ; i ++){
int data = random.nextInt(datas.length);
datas[i] = data;
}
//初始化10万个要查询的数字
for(int i = 0; i < params.length; i ++){
int param = random.nextInt(params.length);
params[i] = param;
}
long begin = System.currentTimeMillis();
// 原始数组转map
for (int i = 0; i < datas.length ; i++){
Integer index = map.get(datas[i]);
if(index == null){
map.put(datas[i], i);
}
else if(i < index)
{
map.put(datas[i], i);
}
}
long end = System.currentTimeMillis();
System.out.println("int 数组 转map花了" + (end -begin) + "毫秒");
}
public void testLong(){
long begin = System.currentTimeMillis();
int exist = 0; //下标找到了
int noteexist = 0; //下标没找到
for (int i = 0; i < params.length ; i++){
Integer index = map.get(params[i]);
if(index == null){
noteexist++;
}
else if(i < index)
{
exist++;
}
}
long end = System.currentTimeMillis();
System.out.println("下标找到了:" + exist +"个");
System.out.println("下标没找到:" + noteexist + "个");
System.out.println("查询耗时：" + (end - begin) + "毫秒");
}
public static void main(String[] args){
dataTest dt = new dataTest();
dt.testLong();
}
}