2021-05-10列表、数组、集合的转换及迭代器

最新推荐文章于 2025-06-20 21:44:12 发布

原创最新推荐文章于 2025-06-20 21:44:12 发布 · 175 阅读

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本文介绍了Java中如何高效地进行集合操作，包括数组与集合之间的转换、集合的排序、反转及混洗等技巧，并展示了如何利用迭代器来优化遍历过程。

#tools
##Demo01

package tools;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

/**
 * @qvthor liuwenzheng
 * @date 2021/5/8 11:25
 */
public class Demo01 {
    public static void main(String[] args) {
        String str[] = {"中文","计算机","ABC","123","qq@.com"};
        List<String>list = Arrays.asList(str);  //Arrays.aslist()  将数组转换成列表
        System.out.println(list);

        //2将数组转换成数组列表（链式列表） new ArrayList<>(Arrays.asList(str));
        ArrayList<String> alist = new ArrayList<>(Arrays.asList(str));
        LinkedList<String>llist = new LinkedList<>(Arrays.asList(str));
        alist.remove("abc");
        llist.add("123123");
        System.out.println(alist);
        System.out.println(llist);
        //3将列表转换成数组      列表.toArray();
        Object[] str2 = alist.toArray();
        for (Object o :str2){
            System.out.println(o+"");
        }
        System.out.println();
        Object[] str3 = llist.toArray();
            for (Object o : str3){
                System.out.println(o + "");
            }
        }
    }

##Demo02

package tools;

import java.util.Arrays;
import java.util.HashSet;

/**
 * @qvthor liuwenzheng
 * @date 2021/5/10 8:13
 */
public class Demo02 {
    public static void main(String[] args) {
        String str[] = {"计算机","中文","abc","qq@qq.com","abc"};
        //将数组转换为集合
        HashSet<String>set = new HashSet<>(Arrays.asList());
        System.out.println(set);
        //将集合转换为数组
        Object[]arr = set.toArray();
        for (Object a : arr){
            System.out.println(a);
        }
    }
}

##Demo03

package tools;

import java.util.*;

/**
 * @qvthor liuwenzheng
 * @date 2021/5/10 8:22
 */
public class Demo03 {
    public static void main(String[] args) {
        HashMap<String,String> map = new HashMap<>();
        map.put("1","a");
        map.put("2","b");
        map.put("3","c");
        // 1 keySet()
        Set<String>keys = map.keySet();
        System.out.println(keys);
        // 2 把set转换为其他集合
        ArrayList<String> alistkeys = new ArrayList<>(keys);
        System.out.println(alistkeys);
        // 3去除map的值 value
        Collection<String> v = map.values();
        System.out.println(v);
        // 4 把值转换成其他的集合 new LinkedList<>(v);
        LinkedList<String> llistv = new LinkedList<>(v);
        System.out.println(llistv);

    }

}

##Demo04迭代迭器


 package tools;

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;

/**
 * @qvthor liuwenzheng
 * @date 2021/5/10 9:08
 */
public class Demo04 {
    public static void main(String[] args) {
        HashMap<Integer,String> map = new HashMap<>();
        for (int i = 0 ; i < 20000; i++){
            map.put(i , "haha" + i);
        }
        // 迭代器 减少资源占用
        Iterator<Map.Entry<Integer, String>> iterator = map.entrySet().iterator();
        while (iterator.hasNext()){
            Map.Entry<Integer, String> next = iterator.next();
            System.out.println(next);
        }
    }
}

##Demo05

package tools;

import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;

/**
 * @qvthor liuwenzheng
 * @date 2021/5/10 9:38
 */
public class Demo05 {
    public static void main(String[] args) {
        ArrayList nums = new ArrayList<>();
        nums.add(1);
        nums.add(3);
        nums.add(4);
        nums.add(5);
        nums.add(2);
        // 1 排序
        Collections.sort(nums);
        System.out.println(nums);
        // 2 反转
        Collections.reverse(nums);
        System.out.println(nums);
        // 3 混洗
        Collections.shuffle(nums);
        System.out.println(nums);
        // 4 求最大小值
        System.out.println(Collections.max(nums));
        System.out.println(Collections.min(nums));
        // 5 替换
        Collections.replaceAll(nums,0,100);
        // 6 求频率
        System.out.println(Collections.frequency(nums,100));
        // 7 二分搜索 ， 使用的条件是集合是有序的
        Collections.sort(nums);
        System.out.println(nums);
        int i = Collections.binarySearch(nums,3);
        System.out.println("位置" + i);


    }
}