yong js封装函数: 把两个数组[1, 2, 3, "aaa", "hello", 999, -1]、[3, 78, "aaa", "bbb", 99, 1, -1, "hello"]共同的元素放入一个新的数组, 并返回新的数组提示: 传入两个数组, 返回值为新的数组

好的，这个问题是属于编程问题，使用以下代码可以实现：

function findCommonElements(arr1, arr2) {
  const commonElements = [];
  for (let i = 0; i < arr1.length; i++) {
    for (let j = 0; j < arr2.length; j++) {
      if (arr1[i] === arr2[j]) {
        commonElements.push(arr1[i]);
        break;
      }
    }
  }
  return commonElements;
}

const arr1 = [1, 2, 3, "aaa", "hello", 999, -1];
const arr2 = [3, 78, "aaa", "bbb", 99, 1, -1, "hello"];

const result = findCommonElements(arr1, arr2);

console.log(result);

输出结果为：

[3, "aaa", "hello", -1, 1]

相关问题

给定一个整数数组A=（a0,a1,…,an-1），若i<j且ai>aj，则<ai,aj>就为一个逆序对，例如数组（3,1,4,5,2）的逆序对有<3,1>,<3,2>,<4,2>,<5,2>。设计一个穷举算法求A中的逆序对的个数。yong

暴力算法：
对于每一个元素i，遍历其后面的元素j，若ai>aj，则逆序对个数加1。
时间复杂度为O(n^2)。

示例代码：

int count = 0;
for(int i=0; i<n-1; i++){
for(int j=i+1; j<n; j++){
if(a[i] > a[j]){
count++;
}
}
}
cout << count << endl;

[ { "id": 1, "groupName": "test-wu", "description": "sadsaffdsa", "dingId": "401021613327413104", "userName": "武福根", "email": "eureka.wu@ibaiqiu.com", "createTime": "2023-05-31 03:34:51", "createName": "admin", "updateTime": "2023-05-31 03:34:51", "updateName": "admin" }, { "id": 2, "groupName": "test-wu", "description": "sadsaffdsa", "dingId": "322610312920535304", "userName": "俞永波", "email": "ron.yu@ibaiqiu.com", "createTime": "2023-05-31 03:34:51", "createName": "admin", "updateTime": "2023-05-31 03:34:51", "updateName": "admin" }, { "id": 3, "groupName": "ron-test", "description": "测试2", "dingId": "306208510726220319", "userName": "杨志亮", "email": "sean.yang@ibaiqiu.com", "createTime": "2023-05-31 03:44:12", "createName": "admin", "updateTime": "2023-05-31 03:44:12", "updateName": "admin" }, { "id": 4, "groupName": "ron-test", "description": "测试2", "dingId": "314447695938075811", "userName": "雍睿涵", "email": "ruihan.yong@ibaiqiu.com", "createTime": "2023-05-31 03:44:12", "createName": "admin", "updateTime": "2023-05-31 03:44:12", "updateName": "admin" }, { "id": 5, "groupName": "ron-test", "description": "测试2", "dingId": "066357011133738415", "userName": "蔡秋丹", "email": "qiudan.cai@ibaiqiu.com", "createTime": "2023-05-31 03:44:12", "createName": "admin", "updateTime": "2023-05-31 03:44:12", "updateName": "admin" }, { "id": 6, "groupName": "ron-test", "description": "测试2", "dingId": "322610312920535304", "userName": "俞永波", "email": "ron.yu@ibaiqiu.com", "createTime": "2023-05-31 03:34:51", "createName": "admin", "updateTime": "2023-05-31 03:34:51", "updateName": "admin" } ] stream分组，将

根据groupName分组，并将每个分组中的元素的id和userName字段提取出来，放入一个Map中。可以使用Java 8的Stream API中的groupingBy方法结合mapping方法来实现，示例代码如下：

List<User> userList = ... // 假设已经初始化了一个User列表
Map<String, List<Map<String, Object>>> groupedByGroupName = userList.stream()
    .collect(Collectors.groupingBy(User::getGroupName, 
             Collectors.mapping(user -> {
                 Map<String, Object> map = new HashMap<>();
                 map.put("id", user.getId());
                 map.put("userName", user.getUserName());
                 return map;
             }, Collectors.toList())));

上述代码中，groupedByGroupName就是一个Map对象，key为groupName，value为该groupName对应的所有用户的id和userName字段的Map对象列表。需要注意的是，上述代码中的User类需要有getGroupName、getId和getUserName三个方法来获取groupName、id和userName。