比较两组，以确定第一组是否是其他集合的子集

Question

我正在制作一个android应用程序来选择一个基于规格的产品，并发送带有产品代码的sms。我有大量的不同规格的产品清单。我需要比较一下。23规格选择一种产品。我知道我可以使用多个if语句或嵌套的ifs，但我想知道是否有更好的方法来实现相同的功能。我只能选择产品，如果所有23个规格是完全相同的。所有规范都存储在变量中，如M1到M49，并根据产品代码进行映射。

SMS部分已经完成了，但我很困惑地比较这么多的变量。

任何指示都会受到高度赞赏。

Answer 1

您可以使用JAVA的retainAll函数

class Check
{
static String s[] = { "q","w","e","r","t","y","u","i","o","p","a",
               "mmmmm","d","f","g","h","j","k","l","z","x","c","v" };

static String s1[] = {"b","n","m","er","re","ew","j","k","x","qq","ww","ee","c","v","t","y","d","f","g","h",
               "u","rr","yyy","uuu","ttt","trew","rerer","q","w","vv","gg","hh","tt","tr","ww","e","i"
               ,"eww","qwe","wer","o","p","a","s","ds","cv","r","l","z"};

public static void main(String arags[])
{
   Set<String> originalset = new HashSet<String>(Arrays.asList(s1));
   Set<String> testset = new HashSet<String>(Arrays.asList(s));        
   originalset.retainAll(testset);
    if(originalset.size()==23)
    {
        System.out.println("Sucess 23 matches found");
    }
    else
    {
          System.out.println("Fail 23 matches not found");
    }

}
}

Answer 2

步骤：

1. Put both of them into a HashSet. A HashSet searches faster and doesn't allow duplicate elements.
2. Find if first hashset is a subset of second hashset.


import java.util.Iterator;


public class Sort {
    public static void main(String[] args) {
        String s[] = { "q","q","q","r","t","y","u","i","o","p","a",
                   "m","d","f","g","h","j","k","l","z","x","c","v" };

        String s1[] = {"b","n","m","q","q","er","re","ew","j","k","x","qq","ww","ee","c","v","t","y","d","f","g","h",
                   "u","rr","yyy","uuu","ttt","trew","rerer","q","w","vv","gg","hh","tt","tr","ww","e","i"
                   ,"eww","qwe","wer","o","p","a","s","ds","cv","r","l","z"};

       /* System.out.println("Length of s : " + s.length);
        System.out.println("Length of s1 : " + s1.length);
        Arrays.sort(s);
        Arrays.sort(s1);

        display(s);
        display(s1);
        int j=0;
        int count=0;
        for(int i=0;i<s1.length;i++) {
            if(s[j].equals(s1[i])) {

                count++;
                j++;
                if(j==23) break;
            }
        }
        if(j==23) System.out.println("23 matches found");
        else System.out.println("23 matches not found");
        */

        HashSet<String> hs1 = createSet(s);
        HashSet<String> hs2 = createSet(s1);
        System.out.println(h1SubsetOfh2(hs1, hs2));
    }

    public static HashSet<String> createSet(String[] sArr) {
        HashSet<String> hs = new HashSet<String>();
        for(int i=0;i<sArr.length;i++) {
            hs.add(sArr[i]);
        }
        return hs;
    }

    public static boolean h1SubsetOfh2(HashSet<String> hs1,HashSet<String> hs2) {
        Iterator<String> it = hs1.iterator();

        while(it.hasNext()) {
            if(!hs2.contains(it.next()))
                return false;
        }
        return true;
    }


}

Answer 3

import java.util.Arrays;


public class Sort {
    public static void main(String[] args) {
        String s[] = { "q","w","e","r","t","y","u","i","o","p","a",
                   "mmmmm","d","f","g","h","j","k","l","z","x","c","v" };

    String s1[] = {"b","n","m","er","re","ew","j","k","x","qq","ww","ee","c","v","t","y","d","f","g","h",
                   "u","rr","yyy","uuu","ttt","trew","rerer","q","w","vv","gg","hh","tt","tr","ww","e","i"
                   ,"eww","qwe","wer","o","p","a","s","ds","cv","r","l","z"};

        System.out.println("Length of s : " + s.length);
        System.out.println("Length of s1 : " + s1.length);
        Arrays.sort(s);
        Arrays.sort(s1);

        display(s);
        display(s1);
        int j=0;
        int count=0;
        for(int i=0;i<s1.length;i++) {
            if(s[j].equals(s1[i])) {

                count++;
                j++;
                if(j==23) break;
            }
        }
        if(j==23) System.out.println("23 matches found");
        else System.out.println("23 matches not found");


    }
    public static void display(String a[]) {

        for(int i=0;i<a.length;i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println();
    }



}

试试这个。