数组编程面试题：补全合并对象数组的JavaScript代码

嘿，我来帮你搞定这个对象合并的需求！咱们先理清楚：有两个数组，drunk_ppl存着饮酒者的基本信息，smoker存着吸烟者的吸烟习惯，要把名字相同的对象所有属性合并到一起，最终放到drunkard数组里对吧？

原始代码片段 #

你给出的初始代码是这样的：

drunk_ppl = [ {name: 'Mark', age: 20, alcohol_type: 'Whiskey'}, {name: 'Jon', age: 25, alcohol_type: 'Rum'}, {name: 'April', age: 22, alcohol_type: 'Gin'}, {name: 'Simon', age: 50, alcohol_type: 'Vodka'} ];
smoker = [ {name: 'Mark', cig_a_day: 20, brand: 'Marlboro'}, {name: 'Jon', cig_a_day: 15, brand: 'Camel'}, {name: 'April', cig_a_day: 12, brand: 'Newport'}, {name: 'Simon', cig_a_day: 5, brand: 'Dunhill'} ];
var drunkard = []; //创建空数组
for(var i = 0; i < drunk_ppl.length; i++...

补全后的基础实现（嵌套循环） #

先给你补全最直观的嵌套循环写法，容易理解：

drunk_ppl = [ {name: 'Mark', age: 20, alcohol_type: 'Whiskey'}, {name: 'Jon', age: 25, alcohol_type: 'Rum'}, {name: 'April', age: 22, alcohol_type: 'Gin'}, {name: 'Simon', age: 50, alcohol_type: 'Vodka'} ];
smoker = [ {name: 'Mark', cig_a_day: 20, brand: 'Marlboro'}, {name: 'Jon', cig_a_day: 15, brand: 'Camel'}, {name: 'April', cig_a_day: 12, brand: 'Newport'}, {name: 'Simon', cig_a_day: 5, brand: 'Dunhill'} ];
var drunkard = []; //创建空数组

for(var i = 0; i < drunk_ppl.length; i++) {
  // 先拿当前的饮酒者对象
  const currentDrunk = drunk_ppl[i];
  // 遍历吸烟者数组找同名的
  for(var j = 0; j < smoker.length; j++) {
    const currentSmoker = smoker[j];
    if(currentDrunk.name === currentSmoker.name) {
      // 用展开运算符合并两个对象的所有属性
      const mergedObj = {...currentDrunk, ...currentSmoker};
      // 也可以用Object.assign：const mergedObj = Object.assign({}, currentDrunk, currentSmoker);
      drunkard.push(mergedObj);
      // 找到匹配的就跳出内层循环，别白跑啦
      break;
    }
  }
}

// 打印看看结果
console.log(drunkard);

更高效的优化写法 #

如果数组里的元素特别多，嵌套循环效率就有点低了。咱们可以先把smoker转成一个以名字为键的映射表，这样找同名对象就不用一遍遍遍历了：

drunk_ppl = [ {name: 'Mark', age: 20, alcohol_type: 'Whiskey'}, {name: 'Jon', age: 25, alcohol_type: 'Rum'}, {name: 'April', age: 22, alcohol_type: 'Gin'}, {name: 'Simon', age: 50, alcohol_type: 'Vodka'} ];
smoker = [ {name: 'Mark', cig_a_day: 20, brand: 'Marlboro'}, {name: 'Jon', cig_a_day: 15, brand: 'Camel'}, {name: 'April', cig_a_day: 12, brand: 'Newport'}, {name: 'Simon', cig_a_day: 5, brand: 'Dunhill'} ];
var drunkard = [];

// 用reduce把smoker转成{名字: 吸烟者对象}的映射
const smokerMap = smoker.reduce((map, item) => {
  map[item.name] = item;
  return map;
}, {});

// 遍历饮酒者数组，直接通过名字取对应的吸烟信息合并
drunk_ppl.forEach(drunk => {
  const smokerInfo = smokerMap[drunk.name];
  if(smokerInfo) {
    drunkard.push({...drunk, ...smokerInfo});
  }
});

console.log(drunkard);

小提示 #

• 合并对象时，如果两个对象有同名属性（比如这里两个数组的对象都有name），后面的对象属性会覆盖前面的。不过在这个场景下name值是一样的，所以完全没问题。
• 优化写法的时间复杂度是O(n)，比嵌套循环的O(n*m)快很多，数据量大的时候优势明显。

内容的提问来源于stack exchange，提问作者Akash Goel