如何在 JavaScript 中对两个对象数组执行内连接？

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'},
]

我想对这两个数组进行内部联接a and b，并创建第三个数组，如下所示（如果位置属性不存在，则它变为 null）：

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'},
}]

我的做法：

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.
        });
    });
}

但时间复杂度是O(N^2)。我怎样才能做到O(N)？我的朋友告诉我我们可以使用减速器Object.assign.

我无法弄清楚这一点。请帮忙。

我不知道怎么办reduce在这里会有所帮助，但你可以使用Map到 完成相同的任务O(n):

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; }

Because Map访问和更新是O(1)（平均而言 - 由于哈希 碰撞和重新散列，它可以更长），这使得O(n+m) (where n and m的长度是a and b分别;你的天真的解决方案 给将是O(n*m)使用相同的含义n and m).