第 30 题：请把俩个数组 [A1, A2, B1, B2, C1, C2, D1, D2] 和 [A, B, C, D]，合并为 [A1, A2, A, B1, B2, B, C1, C2, C, D1, D2, D]。

[jefferyE]

function concatArr (arr1, arr2) {
   const arr = [...arr1];
  let currIndex = 0;
 for (let i = 0; i < arr2.length; i++) {
    const RE = new RegExp(arr2[i])
    while(currIndex < arr.length) {
      ++currIndex
      if (!RE.test(arr[currIndex])) {
         arr.splice(currIndex, 0, a2[i])
         break;
       }
     }
   }
  return arr
 }
 var a1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'] 
 var a2 = ['A', 'B', 'C', 'D']
 const arr = concatArr(a1, a2)
 console.log(a1) // ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'] 
 console.log(a2) // ['A', 'B', 'C', 'D']
 console.log(arr) // ['A1', 'A2', 'A', B1', 'B2', 'B', C1', 'C2', 'C', D1', 'D2', 'D'] 

以上是我个人想法，有更好方法的欢迎讨论

[GitHdu]

const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
const arr2 = ['A', 'B', 'C', 'D']
const ret = []
let tmp = arr2[0]
let j = 0
for (let i=0;i<arr1.length;i++) {
  if (tmp === arr1[i].charAt(0)){
    ret.push(arr1[i])
  }else {
    ret.push(tmp)
    ret.push(arr1[i])
    tmp=arr2[++j]
  }
   if(i===arr1.length-1){
      ret.push(tmp)
    }
}
console.log(ret)

[atheist1]

var a = ['A1','A2','B1','B2','C1','C2','D1','D2']
var b = ['A','B','C','D']
// 对需要排序的数字和位置的临时存储
var mapped = a.concat(b).map(function(el, i) {
  return { index: i, value: /\D$/.test(el) ? (el + 4) : el };
})
mapped.sort(function(a, b) {
  return +(a.value > b.value) || +(a.value === b.value) - 1;
});
var result = mapped.map(function(el){
  return a.concat(b)[el.index];
});

利用mdn对sort映射改善排序的方法进行的处理，不过对数组进行了多次处理，感觉方法不太好

[JJL-SH]

let a1 =  ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
let a2 = ['A', 'B', 'C', 'D'].map((item) => {
  return item + 3
})

let a3 = [...a1, ...a2].sort().map((item) => {
  if(item.includes('3')){
    return item.split('')[0]
  }
  return item
})

[gaomin]

var arr1 = ["A1", "A2", "B1", "B2", "C1", "C2", "D1", "D2"]
var arr2 = ["A", "B", "C", "D"]
var arr3 = arr1.concat(arr2);
arr3.sort().sort(function(a,b){
   if (a.charAt(0) == b.charAt(0) && a.length > b.length){
       return -1
   }
	
})```

[ESnail]

var arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
var arr2 = ['A', 'B','C', 'D'];

function fn (arr1, arr2) {
let arr3 = [...arr1];
let index = -1;
arr2.forEach((v, i) => {
index = index + 3;
arr3.splice(index, 0, v);
});
return arr3;
}

console.log(fn(arr1, arr2)); // [ 'A1', 'A2', 'A', 'B1', 'B2', 'B', 'C1', 'C2', 'C', 'D1', 'D2', 'D' ]
// -1 + 3 = 2
// 2 + 3 = 5
// 5 + 3 = 8
// 8 + 3 = 11
// 首先想到了用concat+sort，但结果不对。后来想观察了一下，用splice插入，因为插入位置是固定的。

[veaba]

看一下我这个可以嘛

var a1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
var a2 =['A','B','C','D'];
var j=-1;
var arr=[]
for(let i=0;i<a1.length;i++){
    if(i%2 ===0){
	j++
	arr=arr.concat((a1.slice(i,i+2)).concat(a2[j]))
    }
}
console.log(arr)

——————————————————————
截图

[acmu]

我想问下，这题想考的是哪方面的知识？

[DraCod]

var arrOne =["A1", "A2", "B1", "B2", "C1", "C2", "D1", "D2"]；
var arrTwo = ['A', 'B', 'C', 'D'];

for (let i = 0; i < arrTwo.length; i++) {
let re = new RegExp(arrTwo[i], 'g');
for (let x = arrOne.length; x > 0; x--) {
if(re.test(arrOne[x])){
arrOne.splice(x+1,0,arrTwo[i])
}
}
}
console.log(arrOne);

这样是否可以呢？

[dorseysen]

[Rashomon511]

let arrA = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
let arrB = ['A', 'B', 'C', 'D',];
function combine(a, b) {
  while (b.length){
    let str =  b.shift();
    let indexNum = 0;
    a.forEach((item,index) => {
      if(item.indexOf(str) !== -1){
        indexNum = index
      }
    })
    a.splice(indexNum + 1, 0, str)
  }
  return a;
}

[tyosssss]

var arr1 = ["A1", "A2", "B1", "B2", "C1", "C2", "D1", "D2"];
var arr2 = ["A", "B", "C", "D"];

arr2.forEach((it, index) => {
  arr1.splice((index + 1) * 2 + index, 0, it);
});

console.log(arr1);

[blockmood]

我想问下，这题想考的是哪方面的知识？

假设有一种情况，让你在一个列表中插入一个广告，不光是数组，对象依然有这种需求，这道题其实就是平常经常需要用到的一个小功能。

[Mizxinp]

const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
const arr2 = ['A', 'B', 'C', 'D']
const arr = [...arr1,...arr2]
let targetArr = [];
arr2.forEach(item => {
  arr.forEach(ele=>{
      if(ele.includes(item)){
	targetArr.push(ele)
       }
   })
});	
console.log('targetArr',targetArr);

[bigbigbo]

如果只是单纯解这道题的话，我这样做：

const res = ["A", "B", "C", "D"].reduce(
  (memo, item) => {
    const tmp = [...memo].reverse();
    const idx = memo.length - tmp.findIndex(i => i.startsWith(item)) - 1;

    return [...memo.slice(0, idx + 1), item, ...memo.slice(idx + 1)];
  },
  ["A1", "A2", "B1", "B2", "C1", "C2", "D1", "D2"]
);

这样即使是["A1", "A2", "A3", "B1", "B2", "B3", "C1", "C2", "D1", "D2"]等等也没问题啦

[Moriarty02]

const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
const arr2 = ['A', 'B', 'C', 'D']
function combineArr(a1, a2) {
  const ret = a2.map((item, index) => {
    return [a1[index * 2], a1[index * 2 + 1], item]
  })
  return ret.toString().split(',')
}
console.log(combineArr(arr1, arr2))

2个参考点

1. arr1和arr2本身就是有序的，A1的index和A的index成2倍关系
2. 可以使用Array的toString方法将数组拍平