javascript - 如何在 JavaScript 中对两个对象数组执行内部连接？

Question

我有两个对象数组:

var a = [
  {id: 4, name: 'Greg'},
  {id: 1, name: 'David'},
  {id: 2, name: 'John'},
  {id: 3, name: 'Matt'},
]

var b = [
  {id: 5, name: 'Mathew', position: '1'},
  {id: 6, name: 'Gracia', position: '2'},
  {id: 2, name: 'John', position: '2'},
  {id: 3, name: 'Matt', position: '2'},
]

我想对这两个数组 a 和 b 进行内部连接，并像这样创建第三个数组(如果 position 属性不存在，那么它变为空):

var result = [{
  {id: 4, name: 'Greg', position: null},
  {id: 1, name: 'David', position: null},
  {id: 5, name: 'Mathew', position: '1'},
  {id: 6, name: 'Gracia', position: '2'},
  {id: 2, name: 'John', position: '2'},
  {id: 3, name: 'Matt', position: '2'},
}]

我的方法:

function innerJoinAB(a,b) {
    a.forEach(function(obj, index) {
        // Search through objects in first loop
        b.forEach(function(obj2,i2){
        // Find objects in 2nd loop
        // if obj1 is present in obj2 then push to result.
        });
    });
}

但是时间复杂度是O(N^2)。如何在 O(N) 中完成？我的 friend 告诉我，我们可以使用 reducer 和 Object.assign。

我无法弄清楚这一点。请帮忙。

Answer 1

我不知道 reduce 在这里有什么帮助，但你可以使用 Map 来 在 O(n) 中完成相同的任务:

const a = [
  {id: 4, name: 'Greg'},
  {id: 1, name: 'David'},
  {id: 2, name: 'John'},
  {id: 3, name: 'Matt'}];

const b = [
  {id: 5, name: 'Mathew', position: '1'},
  {id: 6, name: 'Gracia', position: '2'},
  {id: 2, name: 'John', position: '2'},
  {id: 3, name: 'Matt', position: '2'}];

var m = new Map();
// Insert all entries keyed by ID into the Map, filling in placeholder
// 'position' since the Array 'a' lacks 'position' entirely:
a.forEach(function(x) { x.position = null; m.set(x.id, x); });

// For values in 'b', insert them if missing, otherwise, update existing values:
b.forEach(function(x) {
    var existing = m.get(x.id);
    if (existing === undefined)
        m.set(x.id, x);
    else
        Object.assign(existing, x);
});

// Extract resulting combined objects from the Map as an Array
var result = Array.from(m.values());

console.log(JSON.stringify(result));

.as-console-wrapper { max-height: 100% !important; top: 0; }

因为 Map 访问和更新是 O(1)(平均 - 因为散列 冲突和重新散列，它可以更长)，这使得 O(n+m) (其中 n 和m分别是a和b的长度；你天真的解决方案 gave 将是 O(n*m)，对 n 和 m 使用相同的含义。