抢分，算法问题

[extraleo]

有2个数组A和B（都是数字）
怎么样最快的找到A 和B相同的部分？

[zazxzx]

楼上好强 学习了

[malong_01223]

五楼的不好用！

     */
    public static void main(String[] args) {
        String[] a = {"1","2","3","4","5","3","4"};
        String[] b = {"3","4","5","6"};
        Set s = new HashSet(Arrays.asList(a));
        s.retainAll(Arrays.asList(b));
        System.out.println(s);
    }

结果是[3, 5, 4]
我们希望借结果是[3, 4, 5]

[lionest]

ＬＳ的不错！

[醉面韦陀]

比较笨的方法：

import java.util.ArrayList;
import java.util.List;

/*
 * 取数组相同部分
 */
public class CollectionTest {
	
	public static void main(String[] args) {
		
		int[] num_one={1,2,3,5,4,6};
		int[] num_two={2,5,36,4,6};
		someNum(num_one,num_two);
	}

	@SuppressWarnings("unchecked")
	public static void someNum(int[] num_one, int[] num_two) {
		
		List list=new ArrayList();
		for(int num1 : num_one)
		{
			for(int num2 : num_two)
			{
				if(num1==num2)
				{
					list.add(num1);
				}
			}
		}
		showList(list);
	}

	public static void showList(List list) {
		
		for(Object num : list)
		{
			System.out.println(num.toString());
		}
		
	}
}

[mxx123]

用循环

[czrjsj]

字符串中的相同部分可能不止一个
最好说明详细一点

[zyclc]

5楼真有趣

[zhitaorao]

问题解决了就好,何必在意哪一个方法呢?

[linxi1151]

String[] a = {"1","2","3","4","5","1","2"};
String[] b = {"3","4","5","6","1","2"};

象这样的字符窜算几部分相同的？运行出[3, 2, 1, 5, 4]这样的结果能行吗

[zhangbaokun]

我晕死了用TreeSet就排序了，汗

import java.util.*;

public class Test
{
	public final static String POEM="This this this one. a one sb SD sd";

	public static void main(String[] args) {
        String[] a = {"1","2","3","4","5","3","4"};
        String[] b = {"3","4","5","6"};
        Set s = new TreeSet(Arrays.asList(a));
        s.retainAll(Arrays.asList(b));
        System.out.println(s);
    }

}

[zhuode_001]

关注，顺便接分

[zhuode_001]

public class ss extends S {
	

	public ss(double rrr){
		super(rrr);
		
	}
	
	public double getArea(){
		return 3.14*r;
		
	}
	
	public static void main(String[] args){
		ss sss = new ss(10);
		System.out.println(""+sss.getArea());
	}
}

[jinxfei]

指的是连续相同的子数组，还是所有相同的元素？

[shan1119]

循环.

[karlpan01]

又是这样的帖子!

public class Test{
    /**
     * 获取两个整型数组之间的重复元素集合
     * @param array1 数组参数1
     * @param array2 数组参数2
     * @return
     */
    public List findSame(int array1[],int array2[]){
        List result=new ArrayList();//重复元素结果集合
        HashMap hashMap=new HashMap();//利用hashmap来寻找重复元素
        for(int i=0;i<array1.length;i++){//将第一个数组加入hashmap
            String temp=array1[i]+"";
            hashMap.put(temp,temp);
        }
        for(int i=0;i<array2.length;i++){//遍历第二个数组
            String temp=array2[i]+"";
            if(hashMap.get(temp)!=null){//在已经存在第一个数组所有元素的hashmap里寻找第二数组里的元素
                result.add(array2[i]);//将重复出现的元素加入结果集合
            }
        }
        return result;
    }

    public static void main(String args[]){
        long timeBegin=System.currentTimeMillis();
        int   a[]   =   {1,   6,   2,   8,   5,   8,   6,   9,   0}; 
        int   b[]   =   {4,   5,   4,   8,   7,   6,   2,   0}; 
                //获取重复元素集合
        List list=new Test().findSame(a, b);
        //遍历输出重复元素
        for(int i=0;i<list.size();i++){
            System.out.println(list.get(i));
        }
        }
}

这是那次一个大哥写的!

[karlpan01]

那个大哥叫lip009!
http://topic.csdn.net/u/20071113/18/9fa94e23-df29-432f-8e21-b9498cb00a0d.html
再把那个帖子给你,你自己看吧!

[zhangbaokun]

操作不了指针再快能快到哪去，要快就不在这弄了

[he156294775]

数据结构里有个字符串的匹配算法,回去找本就能看明白.

[wangjemmy]

虽然不配合该版,不过提一下C# 3.0

 string[] a = {"1","2","3","4","5","3","4"};
 string[] b = {"3","4","5","6"};
 string[] c = a.Intersect(b).ToArray();
 foreach(string s in c)
    Console.WriteLine(s)

或者:
 string[] a = {"1","2","3","4","5","3","4"};
 string[] b = {"3","4","5","6"};
 var s=from c in b where a.Contains(c)
       select c;
 foreach(string s in c)
    Console.WriteLine(s)

[小远]

果然都是学JAVA的...这是算法题..为快不是让想方法实现面向对象思想到都不错