在两个数组之间迭代以找到匹配值的最有效方法是什么

Question

我需要通过匹配id数组来查找数组中的对象。id数组可以更长，也可以等于人员数组的长度。我用forEach循环的persons数组和内部使用includes方法来找到匹配的id，但不确定这是不是一个好方法。有没有优化搜索算法的方法？

const ids = [1, 4, 9, 7, 5, 3];
const matchedPersons = [];
const persons = [
{
id: 1,
name: "James"
},
{
id: 2,
name: "Alan"
},
{
id: 3,
name: "Marry"
}
];
persons.forEach((person) => {
if (ids.includes(person.id)) {
matchedPersons.push(person);
}
});
console.log(matchedPersons);

密码箱

Answer 1

您可以使用带有O(1(的Set进行检查。

const
ids = [1, 4, 9, 7, 5, 3],
persons = [{ id: 1, name: "James" }, { id: 2, name: "Alan" }, { id: 3, name: "Marry" }],
idsSet = new Set(ids),
matchedPersons = persons.filter(({ id }) => idsSet.has(id));
console.log(matchedPersons);

Answer 2

您最好使用filter。它做的正是它应该做的：

const ids = [1, 4, 9, 7, 5, 3];
const persons = [
{
id: 1,
name: "James"
},
{
id: 2,
name: "Alan"
},
{
id: 3,
name: "Marry"
}
];
const matchedPersons = persons.filter(({id}) => ids.includes(id))
console.log(matchedPersons)

Answer 3

您可以使用Maphttps://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map/get

const ids = [1, 4, 9, 7, 5, 3];
const matchedPersons = [];
const persons = [
{
id: 1,
name: "James"
},
{
id: 2,
name: "Alan"
},
{
id: 3,
name: "Marry"
}
];
const personsMap = new Map()
persons.forEach((person) => {
personsMap.set(person.id, person)    
});
persons.forEach((person) => {
if (personsMap.has(person.id)) {
matchedPersons.push(personsMap.get(person.id));
}
});
console.log(matchedPersons);