javascript - 如何删除引用数组的额外出现

Question

假设我们有两个这样的数组:

let s1 =  ["A", "B", "C", "D", "B"]  // Note: "B" has been occurred 2 times
let s2 =  ["A", "B", "C", "D", "B", "A", "X", "Y"] // // Note: "B" has been occurred 2 times just like s1 but we have another occurrence for "A" 

我想基于这两个数组创建一个新数组 (let s3)，我们将删除出现次数多于同一元素在数组中出现次数的元素 s1

所以 s3 数组 应该是这样的:

let s3 =  ["A", "B", "C", "D", "B", "X", "Y"] // we just removed "A" cause it occurred more than "A" occurances in s1

请注意，我们需要任何其他内容 ("X", "Y")。我们只想处理额外的事件。

到目前为止，我可以找到像这样重复出现的索引，但我找不到一种方法来比较每次出现以查看它是否是额外的(复杂的):

    let s1 =  ["A", "B", "C", "D", "B"]  
    let s2 =  ["A", "B", "C", "D", "B", "A", "X", "Y"]  

    var s3 = [];  
     
    for (let i = 0; i < s2.length; i++) {
    for (let j = 0; j < s1.length; j++) {
     if (s2[i] === s1[j]) {
         s3.push(i);
         break;
  }
}        
  }     
    console.log(s3)

我的英文很差，不知道能不能解释清楚!

注意:我可以简单地使用 s3 = [...new Set(s2)] 来删除重复的元素，但我想要其他东西。

Answer 1

由于您希望将所有内容保留在 s1 中，但只添加 s2 中尚未存在的项目，因此您可以先设置 s3 > 到 s1。然后遍历 s2，如果 s1 没有 的值，则将其插入 s3。

const s1 = ["A", "B", "C", "D", "B"];
const s2 = ["A", "B", "C", "D", "B", "A", "X", "Y"];

const s3 = [...s1];

s2.forEach(d => {
  if (!s1.includes(d)) s3.push(d);
});

console.log(s3);