java 对重复电话号码进行排除的优化（排序和前后对比）

1.先对10万数据排序；

2.对比前后两条数据 ；

3.筛选出正确数据。

package com.xxx.xxx;

import java.util.ArrayList;
import java.util.Collections;
import java.util.Random;

public class Test2 {
    final static int _capacity = 1000000;
    final static Random rand = new Random(System.currentTimeMillis() + _capacity);
    static ArrayList<String> list = new ArrayList<String>(_capacity);
    static ArrayList<String> newlist = new ArrayList<String>(_capacity);

    public static void main(String[] args) throws InterruptedException {
        long ts = System.currentTimeMillis();
        int modVal = _capacity / 3;
        for (int i = 0; i < _capacity; i++) {
            rand.setSeed(i);
            list.add(Integer.toString(Math.abs(rand.nextInt() % modVal)));
        }
        ts = System.currentTimeMillis() - ts;
        // 生成一个100000条数据的list
        System.out.println("生成时间 ：" + ts);
        test2();
    }

    static void test() {
        newlist.clear();
        int repetition = 0;
        long ts = System.currentTimeMillis();
        
        Collections.sort(list);
        String str = list.get(0);
        int max = list.size();
        for (int i = 1; i < max; i++) {
            if (str.equals(list.get(i))) {
                repetition++;
                continue;
            }
            newlist.add(str);
            str = list.get(i);
        }
        newlist.add(str);

        ts = System.currentTimeMillis() - ts;
        System.out.println("------ 排序检查方法 -------");
        System.out.println("查找时间 ：" + ts);
        System.out.println("重复 ：" + repetition);
        System.out.println("正确 ：" + newlist.size());
    }
    
    
}