取两个String数组的交集

时间:2025-03-05 20:07:26

取两个string数组的交集,首先将第一个数组的值作为key,value为false存储在map集合中;然后将第二个数组的值循环去判断map中key是否存在,存在就讲key对应的value改成true,否则不做变化;
最后,取出map中value为true的key,就是两个数组的交集。

备注:方法一当数组中有重复数据时,map设置key时会把重复的值丢掉,方法二会将两个数组中所有重复值打印出来;

方法一:

1 import org.testng.annotations.Test;
2 import java.util.HashMap;
3 import java.util.LinkedList;
4 import java.util.List;
5 import java.util.Map;
6
7
8 public class test {
9 @Test//测试程序
10 public void test(){
11 String[] arr1 = {"112","wqw","2121"};
12 String[] arr2 = {"112","aad","ewqw"};
13 String[] result=StringIntersection(arr1,arr2);
14 for (String str:result){
15 System.out.printf(str);
16 }
17 }
18 //取两个string数组的交集
19 public String[] StringIntersection(String[] arr1,String[] arr2){
20 Map<String,Boolean> map = new HashMap<String,Boolean>();
21 List<String> list = new LinkedList<String>();
22 //取出str1数组的值存放到map集合中,将值作为key,所以的value都设置为false
23 for (String str1:arr1){
24 if (!map.containsKey(str1)){
25 map.put(str1,Boolean.FALSE);
26 }
27 }
28 //取出str2数组的值循环判断是否有重复的key,如果有就将value设置为true
29 for (String str2:arr2){
30 if (map.containsKey(str2)){
31 map.put(str2,Boolean.TRUE);
32 }
33 }
34 //取出map中所有value为true的key值,存放到list中
35 for (Map.Entry<String,Boolean> entry:map.entrySet()){
36 if (entry.getValue().equals(Boolean.TRUE)){
37 list.add(entry.getKey());
38 }
39 }
40 //声明String数组存储交集
41 String[] result={};
42 return list.toArray(result);
43 }
44 }

方法二:

package com.java8;

import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors; public class StringArrayTest { public static void main(String[] args) {
String[] arr1 = {"112","wqw","2121","112"};
String[] arr2 = {"112","aad","ewqw", "112"};
List list1 = Arrays.asList(arr1); //将数组转化为list
List list2 = Arrays.asList(arr2);
List list = (List) list1.stream().filter(a -> list2.contains(a)).collect(Collectors.toList());
System.out.println(list); //打印出list String[] arr = (String[])list.toArray(new String[list.size()]); //转化为数组
for (String a:arr){
System.out.println(a); //打印出数组中每个元素
}
}
}