一道阿里面试题，求高手解答

atbkw 2018-12-28 08:49:12

一个List放入多个person对象，每个person对象有name,age两个属性，现要求不准遍历List，取出name=张三，age=23的person对象

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110成成 2019-01-10

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数据结构基本算法，递归。

nayi_224 2019-01-10

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感觉这是个脑筋急转弯，虽然说了用List，但是List只是个接口，我完全可以写一个基于hash的list。至于碰撞问题，增大列表长度可以减少碰撞。如果内存无限，hash算法合适，并且数组长度无限，理论上可以解决碰撞。

package test.gt60;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Test69 {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		Person69 p1 = new Person69("p1", 11);
		Person69 p2 = new Person69("p1", 11);
		Person69 p3 = new Person69("p2", 11);
		Person69 p4 = new Person69("p3", 11);
		Person69 p5 = new Person69("p4", 11);
		
		TestHashMapList list = new TestHashMapList();
		
		list.add(p1.toString(), p1);
		list.add(p2.toString(), p2);
		list.add(p3.toString(), p3);
		list.add(p4.toString(), p4);
		list.add(p5.toString(), p5);
		
		System.out.println(list.hashGet("p1_11"));
		
	}

}

class Person69{
	private String name;
	private int age;
	
	public Person69(String name, int age) {
		super();
		this.name = name;
		this.age = age;
	}
	
	@Override
	public String toString() {
		// TODO Auto-generated method stub
		return this.name + "_" + this.age;
	}
	
	public String getName() {
		return name;
	}
	public void setName(String name) {
		this.name = name;
	}
	public int getAge() {
		return age;
	}
	public void setAge(int age) {
		this.age = age;
	}
}

class TestHashMapList {

	private Map<String, List<Person69>> mapValue = new HashMap(2147483);
	
	public void add(String str, Person69 person){
		if(this.mapValue.containsKey(str)){
			this.mapValue.get(str).add(person);
		}else {
			List temp = new ArrayList(16);
			temp.add(person);
			this.mapValue.put(str, temp);
		}
	}
	
	public List<Person69> hashGet(String str){
		return this.mapValue.get(str);
	}
}

stacksoverflow 2019-01-10

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这道题出得有问题，前提条件模糊并且看不出来要考察什么。

nayi_224 2019-01-10

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indexOf还是算了吧

    /**
     * Returns the index of the first occurrence of the specified element
     * in this list, or -1 if this list does not contain the element.
     * More formally, returns the lowest index <tt>i</tt> such that
     * <tt>(o==null ? get(i)==null : o.equals(get(i)))</tt>,
     * or -1 if there is no such index.
     */
    public int indexOf(Object o) {
        if (o == null) {
            for (int i = 0; i < size; i++)
                if (elementData[i]==null)
                    return i;
        } else {
            for (int i = 0; i < size; i++)
                if (o.equals(elementData[i]))
                    return i;
        }
        return -1;
    }

weixin_43191973 2019-01-10

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stream get到了

rickylin86 2019-01-08

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import java.util.ArrayList;


class Person{
	public Person(String name,int age){
		this.name = name;
		this.age = age;
	}


	@Override
	public String toString(){
		return String.format("Person[%s,%d]",name,age);
	}

	@Override
	public boolean equals(Object obj){
		if(!(obj instanceof Person)){
			return false;
		}
		Person person = (Person)obj;
		return person.name.equals(name) && person.age == age;
	}

	private String name;
	private int age;
}

public class Test{
	public static void main(String[] args){
		Person p1 = new Person("李四",28);
		Person p2 = new Person("张三",25);
		Person p3 = new Person("张三",23);
		Person p4 = new Person("王五",24);
		ArrayList<Person> people = new ArrayList<>();
		people.add(p1);
		people.add(p2);
		people.add(p3);
		people.add(p4);

		Person find = new Person("张三",23);

		int index = people.indexOf(find);
		if(index > -1){
			System.out.println("找到:" + people.get(index) + " 对应位置:" + index);
		}else{
			System.out.println("未找到:" + find);
		}
	}
}