将students转换为Map（O（n）），并将其用作reduce的累加器：

new Map(student.map(o => [o.id, { ...o }]))

时间复杂度为O（m）。从累加器（m）中获取student，如果grade或marks不是null，则使用Nullish coalescing assignment ( ??= )将它们分配给student，以避免更新现有属性：

if(grade !== null) student.grade ??= grade;
if(marks !== null) student.marks ??= marks;

复杂度为O（n）+ O（m）=> O（n + m）。

示例（TSPlayground）：

const marks=[{"id":1,"marks":null,"grade":"A"},{"id":1,"marks":90,"grade":null},{"id":1,"marks":90,"grade":"A"},{"id": 2, "marks": 65, "grade":"B"}]
const students =[{"id":1,"name":"john"},{"id": 2, "name": "anna"}]

const result = Array.from(marks.reduce((m, { id, marks, grade }) => {
    const student = m.get(id)

    if(!student) return m

    if(grade !== null) student.grade ??= grade;
    if(marks !== null) student.marks ??= marks;

    return m
  }, new Map(students.map(o => [o.id, { ...o }])))
  .values()
)

console.log(result)

要查找marks和student数组之间的匹配项，可以使用嵌套循环来比较两个数组中每个对象的id属性。但是，为了优化循环以提高效率，您可以使用哈希表将每个student对象的id属性存储为键，并将student数组中的对象索引存储为值。然后，您可以循环遍历marks数组，对于每个对象，使用哈希表检查是否有对应的student对象。时间复杂度为O（n^2），时间复杂度为O（n）。

var marks=[{"id":1,"marks":null,"grade":"A"},{"id":1,"marks":90,"grade":null},{"id":1,"marks":90,"grade":"A"},{"id": 2, "marks": 65, "grade":"B"}]
var students=[{"id":1,"name":"john"},{"id": 2, "name": "anna"}]

// Create a map of student IDs
const studentMap = new Map(students.map((student) => [student.id, student]));

// Loop through marks and find matching students
marks.filter(mark => mark.marks && mark.grade)
     .map(mark => {
       const student = studentMap.get(mark.id);
       if (student) {
         // Execute your code here
         console.log(`${student.name} scored ${mark.marks} and got a grade of ${mark.grade}`);
     }
});