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]。 #39

jefferyE · 2019-03-11T01:39:54Z

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 · 2019-03-11T01:54:15Z

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 · 2019-03-11T01:56:26Z

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 · 2019-03-11T02:07:15Z

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 · 2019-03-11T03:00:03Z

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 · 2019-03-11T06:34:05Z

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 · 2019-03-11T06:46:15Z

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 · 2019-03-11T06:57:07Z

我想问下，这题想考的是哪方面的知识？

DraCod · 2019-03-11T07:35:23Z

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 · 2019-03-11T07:37:47Z

Rashomon511 · 2019-03-11T08:22:14Z

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 · 2019-03-11T10:44:20Z

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 · 2019-03-11T11:13:15Z

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 · 2019-03-11T13:02:28Z

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 · 2019-03-11T15:33:15Z

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 · 2019-03-12T02:14:09Z

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

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

fengT-T · 2019-03-12T07:16:37Z

const matchIndex = str => str.match(/\d+/) || []
const getCharCode = str => str.match(/\w/)[0].charCodeAt()
const result = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
  .concat(['A', 'B', 'C', 'D'])
  .sort((a,b) => {
    const [[aIndex = Infinity], [bIndex = Infinity]] = [matchIndex(a), matchIndex(b)]
    const [aChar, bChar] = [getCharCode(a), getCharCode(b)]
    return aChar === bChar
      ? aIndex - bIndex
      : aChar - bChar
  })
console.log(result)

提一个新思路的版本，从修改sort入手，不依赖数组下标，通用性更强

Lueny-cn · 2019-03-12T08:50:40Z

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

function concatArr(arr1, arr2) {
  let newArr = []
  
  while (arr2.length !== 0) {
    let tag2 = arr2.pop()
    
    newArr.unshift(tag2)
    
    while (arr1.length !== 0) {
      let tag1 = arr1.pop()
      
      if (tag1.includes(tag2)) {
        newArr.unshift(tag1)
      } else {
        arr1.push(tag1)
        break
      }
    }
  }
  return newArr
}

console.log(arr1)
console.log(arr2)
console.log(concatArr(arr1, arr2))

ghost · 2019-03-12T13:00:23Z

其实解法很简单的

let arr1 = ["A1", "A2", "B1", "B2", "C1", "C2", "D1", "D2"];
let arr2 = ["A", "B", "C", "D"];
console.log(
  [...arr1, ...arr2]
    .sort(
      (v2, v1) => (
        v2.codePointAt(0) - v1.codePointAt(0) ||
        v1.length - v2.length ||
        v2.codePointAt(1) - v1.codePointAt(1)
      )
    )
);

jjeejj · 2019-03-12T14:13:33Z

其实解法很简单的

let arr1 = ["A1", "A2", "B1", "B2", "C1", "C2", "D1", "D2"];
let arr2 = ["A", "B", "C", "D"];
console.log([...arr1, ...arr2].sort((v2, v1)=>(v2.codePointAt(0) - v1.codePointAt(0) ? v2.codePointAt(0) - v1.codePointAt(0) : (v1.length - v2.length) || v2.codePointAt(1) - v1.codePointAt(1))));

@liuliangsir 解法是对的，就是你这个函数能不能换行啊，这样长看着多不舒服

BaconZhang · 2019-03-13T06:36:31Z

const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
const arr2 = ['A', 'B', 'C', 'D']
const res = [].concat(...arr2.map(i => arr1.filter(j => j.startsWith(i)).concat(i)))
console.log(res)

cb3570594 · 2019-03-14T01:38:52Z

let arrA = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
let arrB = ['A', 'B', 'C', 'D',];
arrA.map(item => {
  if(item == arrB[0]+2) {return [item,arrB.shift()] } 
  else{return item}
}).flat()

借鉴了前面各位大神的。

yeyan1996 · 2019-03-14T02:31:17Z

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

const func = (arr1, arr2) => arr2.reduce((acc, cur) => [...acc, ...arr1.filter(item => item.startsWith(cur)), cur], [])

WozHuang · 2019-03-14T02:44:28Z

题目测试用例有点少规则太模糊了。。

写一个用sort的吧

const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
const arr2 = ['A', 'B', 'C', 'D'];
const arr3 = arr1.concat(arr2);
const comp = function(a,b){
    const len = Math.max(a.length, b.length);
    for(let i = 0; i < len; i++){
        if(a.charAt(i) === "") return 1;
        if(b.charAt(i) === "") return -1;
        if(a.charAt(i) !== b.charAt(i)){
            return a.charAt(i) > b.charAt(i) ? 1:-1;
        }
    }
    return 0;
}
arr3.sort(comp);
console.log(arr3);

ghost · 2019-03-14T10:33:15Z

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
})

bran-nie · 2019-03-15T06:13:54Z

let a =  ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
let b = ['A', 'B', 'C', 'D']
let arr = a.concat(b).sort()

let tmp = '', r = []
arr.forEach((item, index, a) => {
    if (item.length === 1 && tmp === '') {
	tmp = a[0]
    } else if (item.length === 1) {
	r.push(tmp)
	tmp = item
    } else {
	r.push(item)
    }

    // 这里是将最后获取的单个值，push到最后。
    if (index === a.length -1) {
	r.push(tmp)
    }
})

console.log(r) // ["A1", "A2", "A", "B1", "B2", "B", "C1", "C2", "C", "D1", "D2", "D"]

想法是将合并后的数组sort完，将如A、B替换到A1等的后面。这样的好处是只需要一次遍历。缺点是仅仅针对题目的数据格式，即A1 和 A的字符串长度。

其实一开始想到的是下面这种替换合并后的数组。不过感觉不如新起一个数组

let a =  ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
let b = ['A', 'B', 'C', 'D']
let arr = a.concat(b).sort()
let tmp = ''
arr.forEach((item, index, a) => {    
    if(item.length === 1 && tmp === '') {
        tmp = a.splice(0, 1)
    } else if(item.length === 1) {
        tmp = a.splice(index, 1, ...tmp)
    } else if(index === a.length -1) {
    	a.push(...tmp)
    }
})
console.log(arr) // ["A1", "A2", "A", "B1", "B2", "B", "C1", "C2", "C", "D1", "D2", "D"]

underlineMY · 2019-03-15T07:15:45Z

let arr1 = ['A1', 'A2', 'A3', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
let arr2 = ['A', 'B', 'C', 'D','E'];
let newArr1 = [];
newArr1 = arr2.map((item)=>{
	let arr3 = arr1.filter((value)=> value.startsWith(item));
	arr3.push(item);
	return arr3
})
console.log(newArr1.join(',').split(','));
//["A1", "A2", "A3", "A", "B1", "B2", "B", "C1", "C2", "C", "D1", "D2", "D", "E"]

zhoufeifan · 2019-03-15T08:16:18Z

  const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
  const arr2 = ['A', 'B', 'C', 'D']
  const arr = [...arr1,...arr2]
  arr.sort((a,b)=>{
    // 用字符 : 去做默认占位，因为它 的 ASCII 码大于数字9
    if(a.length > b.length) {
      b = b.padEnd(a.length,':')
    }else {
      a = a.padEnd(b.length,':')
    }
    if(a>b) return 1
    if(a<b) return -1
    return 0
  })
 console.log(arr)

YunShengTeng · 2019-03-15T08:37:19Z

const a = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
const b = ['A', 'B', 'C', 'D'];
// ['A1', 'A2', 'A', 'B1', 'B2', 'B', 'C1', 'C2', 'C', 'D1', 'D2', 'D'];

for (let i = 0; i < b.length; i++) {
  const n = b[i];
  const xx = i + (2 * (i + 1));
  a.splice(xx, 0, n);
}

console.log(a);

DeyaoCai · 2019-03-15T13:55:37Z

console.log(
  function concatArr() {
    return [].concat.apply([], arguments).sort((a, b) => {
      // 如果a 小 在前 则返回 -1
      return a[0] < b[0]
        ? -1
        : a[1]
          ? 1
          : a[1] - b[1];
    });
  }(['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'], ['A', 'B', 'C', 'D'])
);

sunseekers · 2019-03-16T02:44:52Z

//利用字符串的charCodeAt 进行排序，some 只要找到一个符合条件的就不要在进行循环了
var array = ['A1', 'A2','B1', 'B2', 'C1', 'C2', 'D1', 'D2']
var array1 = ['A','B','C','D']
var result = array//重新定义一个新的数组，不影响原来的数组
var array2 = array.join('').split('')
array1.some((x,index)=>{
array2.some((y,key)=>{
if(y.charCodeAt()>x.charCodeAt()){
return result.splice((key/2)+index,0,x)
}
})
})
result.push(array1[array1.length-1])

SceneryCN · 2023-02-20T04:16:53Z

方法比较low但是应该是能实现

const joinArr = (arrA, arrB) => {
  let fisrt = arrA[0].replace(/\d/g, "");
  for (let i = 0; i < arrA.length; i++) {
    if (
      arrA[i].replace(/\d/g, "") !== fisrt &&
      arrB.indexOf(arrA[i].replace(/\d/g, "")) >= 0
    ) {
      arrA.splice(
        i,
        0,
        arrB.find((item) => item === fisrt)
      );
      fisrt = arrA[i + 1].replace(/\d/g, "");
    }
  }
  return arrA;
};

89466598946659 · 2023-02-20T04:17:15Z

您的邮件已收到！O(∩_∩)O！--------Pan

bin-dogami · 2023-02-20T04:17:23Z

这是来自QQ邮箱的假期自动回复邮件。邮件收到，欢迎常联系！。

cxl-blog · 2023-03-09T05:00:49Z

const arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
const arr2 = ['A', 'B', 'C', 'D']
;[...arr1, ...arr2].sort((a, b) => a.charCodeAt()-b.charCodeAt())

// 不确保合并的顺序可以先对长度排序
;[...arr2, ...arr1].sort((a,b) => b.length - a.length).sort((a, b) => a.charCodeAt()-b.charCodeAt())

ScholatLouis · 2023-03-09T05:01:13Z

您的来信已收到，我会尽快回复，谢谢

1242793152 · 2023-03-13T01:38:30Z

可以利用aplisce插入元素的特性，实现一个更简单的方法
let arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
let arr2 = ['A', 'B', 'C', 'D']
function toggle(arr1,arr2){
let index = 2;
arr2.forEach(element => {
arr1.splice(index,0,element)
index += 3
})
console.log(arr1)
return arr1
}
toggle(arr1,arr2)

luke358 · 2023-03-13T02:04:58Z

同首字母比大小，大的先放，不同首字母比大小，小的先放，最后两个把数组全部放完
时间复杂度 O(n),空间复杂度 O(1)

function merge(arr1, arr2) {
  let res = []
  let len1 = arr1.length;
  let len2 = arr2.length;
  let l1 = 0, l2 = 0;
  while (l1 < len1 && l2 < len2) {
    if (arr1[l1].charAt(0) === arr2[l2].charAt(0)) {
      if (arr1[l1] > arr2[l2]) {
        res.push(arr1[l1++])
      } else {
        res.push(arr2[l2++])
      }
    } else {
      if (arr1[l1] < arr2[l2]) {
        res.push(arr1[l1++])
      } else {
        res.push(arr2[l2++])
      }
    }
  }
  while(l1 < len1) {
    res.push(arr1[l1++])
  }
  while(l2 < len2) {
    res.push(arr2[l2++])
  }
  return res;
}

let arr1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
let arr2 = ['A', 'B', 'C', 'D']
merge(arr1, arr2)
merge(arr2, arr1)

sy-l123 · 2023-04-13T01:55:16Z

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

const concatArrFn = () => {
    arrTwo.forEach((e, i) => {
        arrOne.splice((i + 1) * 3 - 1, 0, e);
    });
    return arrOne;
};

ScholatLouis · 2023-04-13T01:55:41Z

您的来信已收到，我会尽快回复，谢谢

bin-dogami · 2023-04-13T01:55:41Z

这是来自QQ邮箱的假期自动回复邮件。邮件收到，欢迎常联系！。

sunchengfeng01 · 2023-06-08T09:56:50Z

const arr1 = ['A1', "A2", "B1", 'B2', 'C1', 'C2', 'D1', 'D2']
const arr2 = ["A", 'B', 'C', 'D']
const arr3 = []
let idnex=0
arr1.forEach((item,index) => {
if (item[0] == arr2[idnex]) {
arr3.push(item)
if (index === arr1.length - 1) {
arr3.push(arr2.pop())
}
} else {
arr3.push(arr2[idnex])
arr3.push(item)

idnex=idnex+1
}

})
console.log(arr3)

bin-dogami · 2023-06-08T09:57:11Z

这是来自QQ邮箱的假期自动回复邮件。邮件收到，欢迎常联系！。

89466598946659 · 2023-06-08T09:57:11Z

您的邮件已收到！O(∩_∩)O！--------Pan

ScholatLouis · 2023-06-08T09:57:14Z

您的来信已收到，我会尽快回复，谢谢

Chenmin926 · 2023-06-17T11:46:54Z

var a1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
var a2 = ['A', 'B', 'C', 'D']

function fn1(a1, a2) {
let result = []
let current = a2.shift()
for (let i = 0; i < a1.length; i++) {
let next = a1[i + 1]
result.push(a1[i])
if ((next && !next.startsWith(current)) || !next) {
result.push(current)
current = a2.shift()
}
}
console.log(result)
}
fn1(a1, a2)

G-WangZH · 2023-09-19T15:37:41Z

var a1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
var a2 = ['A', 'B', 'C', 'D'];

function concatArr(arr1, arr2) {
    const map = new Map();
    const res = [];
    for (let i of arr2) {
        let temp = arr1.filter(j => j.includes(i));
        map.set(i, temp);
    }
    for (let [k, v] of map.entries()) {
        res.push(...v, k);
    }
    return res;
}

concatArr(a1, a2)
// ['A1', 'A2', 'A', 'B1', 'B2', 'B', 'C1', 'C2', 'C', 'D1', 'D2', 'D']

ScholatLouis · 2023-09-19T15:38:07Z

您的来信已收到，我会尽快回复，谢谢

89466598946659 · 2023-09-19T15:38:08Z

您的邮件已收到！O(∩_∩)O！--------Pan

89466598946659 · 2023-10-10T07:29:53Z

您的邮件已收到！O(∩_∩)O！--------Pan

ScholatLouis · 2023-10-10T07:29:53Z

您的来信已收到，我会尽快回复，谢谢

zz201276 · 2023-10-10T07:32:03Z

let arr1 =  ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'] ,arr2 = ['A', 'B', 'C', 'D']
let sortarr = [...arr1, ...arr2].sort((a: any, b: any) => {
    return a.charCodeAt(0) - b.charCodeAt(0)
})

代码很少的方法

Sberm · 2024-01-01T03:23:55Z

凑个热闹!

let a = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2'];
let b =  ['A', 'B', 'C', 'D'];

let r = [...a, ...b].sort((c, d) => {
    if (c[0] < d[0] || (c[0] == c[d] && c.length > d.length)) {
        return -1;
    }
});
console.log(r);

89466598946659 · 2024-01-01T03:24:17Z

您的邮件已收到！O(∩_∩)O！--------Pan

ScholatLouis · 2024-01-01T03:24:17Z

您的来信已收到，我会尽快回复，谢谢

fayeah · 2024-03-07T04:13:35Z

function mergeArr(arr1, arr2) {
    const result = []
    for(let i=1;i < arr1.length;i++) {
        if (arr1[i].substring(0, 1) !== arr1[i-1].substring(0, 1)) {
            result.push(arr1[i-1])
            result.push(arr2.shift())
        } else {
            result.push(arr1[i-1])
        }
        if (i === arr1.length - 1) {
            result.push(arr1[i])
            result.push(arr2.shift())
        }
    }
    return result
}
mergeArr(['A1', 'A2', 'A3', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2', 'E1'], ['A', 'B', 'C', 'D', 'E'])

ScholatLouis · 2024-03-07T04:14:09Z

您的来信已收到，我会尽快回复，谢谢

89466598946659 · 2024-03-07T04:14:12Z

您的邮件已收到！O(∩_∩)O！--------Pan

Archerry · 2024-07-12T06:19:31Z

var a1 = ['A1', 'A2', 'B1', 'B2', 'C1', 'C2', 'D1', 'D2']
var a2 = ['A', 'B', 'C', 'D']

function concatArray(arr1, arr2) {
var resultArr = []
for (const index in arr2) {
const tag = arr2[index]
const targetArr = arr1.filter(item => item.startsWith(tag))
console.log(targetArr)
resultArr.push(...targetArr, arr2[index])
}
return resultArr
}

var resultArr = concatArray(a1, a2)
console.log(resultArr)

ScholatLouis · 2024-07-12T06:20:08Z

您的来信已收到，我会尽快回复，谢谢

89466598946659 · 2024-07-12T06:20:45Z

您的邮件已收到！O(∩_∩)O！--------Pan

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]。 #39

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]。 #39

Comments

jefferyE commented Mar 11, 2019 • edited Loading

GitHdu commented Mar 11, 2019

atheist1 commented Mar 11, 2019 • edited Loading

JJL-SH commented Mar 11, 2019

gaomin commented Mar 11, 2019

ESnail commented Mar 11, 2019

veaba commented Mar 11, 2019 • edited Loading

acmu commented Mar 11, 2019

DraCod commented Mar 11, 2019

dorseysen commented Mar 11, 2019

Rashomon511 commented Mar 11, 2019

tyosssss commented Mar 11, 2019 • edited Loading

blockmood commented Mar 11, 2019 • edited Loading

Mizxinp commented Mar 11, 2019

bigbigbo commented Mar 11, 2019 • edited Loading

Moriarty02 commented Mar 12, 2019 • edited Loading

fengT-T commented Mar 12, 2019 • edited Loading

Lueny-cn commented Mar 12, 2019 • edited Loading

ghost commented Mar 12, 2019 • edited by liuliangsir Loading

jjeejj commented Mar 12, 2019 • edited Loading

BaconZhang commented Mar 13, 2019

cb3570594 commented Mar 14, 2019

yeyan1996 commented Mar 14, 2019 • edited Loading

WozHuang commented Mar 14, 2019

ghost commented Mar 14, 2019

bran-nie commented Mar 15, 2019 • edited Loading

underlineMY commented Mar 15, 2019

zhoufeifan commented Mar 15, 2019 • edited Loading

YunShengTeng commented Mar 15, 2019

DeyaoCai commented Mar 15, 2019 • edited Loading

sunseekers commented Mar 16, 2019

SceneryCN commented Feb 20, 2023 • edited Loading

89466598946659 commented Feb 20, 2023 via email

bin-dogami commented Feb 20, 2023 via email

cxl-blog commented Mar 9, 2023

ScholatLouis commented Mar 9, 2023 via email

1242793152 commented Mar 13, 2023

luke358 commented Mar 13, 2023 • edited Loading

sy-l123 commented Apr 13, 2023 • edited Loading

ScholatLouis commented Apr 13, 2023 via email

bin-dogami commented Apr 13, 2023 via email

sunchengfeng01 commented Jun 8, 2023

bin-dogami commented Jun 8, 2023 via email

89466598946659 commented Jun 8, 2023 via email

ScholatLouis commented Jun 8, 2023 via email

Chenmin926 commented Jun 17, 2023

G-WangZH commented Sep 19, 2023 • edited Loading

ScholatLouis commented Sep 19, 2023 via email

89466598946659 commented Sep 19, 2023 via email

89466598946659 commented Oct 10, 2023 via email

ScholatLouis commented Oct 10, 2023 via email

zz201276 commented Oct 10, 2023

Sberm commented Jan 1, 2024

89466598946659 commented Jan 1, 2024 via email

ScholatLouis commented Jan 1, 2024 via email

fayeah commented Mar 7, 2024 • edited Loading

ScholatLouis commented Mar 7, 2024 via email

89466598946659 commented Mar 7, 2024 via email

Archerry commented Jul 12, 2024

ScholatLouis commented Jul 12, 2024 via email

89466598946659 commented Jul 12, 2024 via email

jefferyE commented Mar 11, 2019 •

edited

Loading

atheist1 commented Mar 11, 2019 •

edited

Loading

veaba commented Mar 11, 2019 •

edited

Loading

tyosssss commented Mar 11, 2019 •

edited

Loading

blockmood commented Mar 11, 2019 •

edited

Loading

bigbigbo commented Mar 11, 2019 •

edited

Loading

Moriarty02 commented Mar 12, 2019 •

edited

Loading

fengT-T commented Mar 12, 2019 •

edited

Loading

Lueny-cn commented Mar 12, 2019 •

edited

Loading

ghost commented Mar 12, 2019 •

edited by liuliangsir

Loading

jjeejj commented Mar 12, 2019 •

edited

Loading

yeyan1996 commented Mar 14, 2019 •

edited

Loading

bran-nie commented Mar 15, 2019 •

edited

Loading

zhoufeifan commented Mar 15, 2019 •

edited

Loading

DeyaoCai commented Mar 15, 2019 •

edited

Loading

SceneryCN commented Feb 20, 2023 •

edited

Loading

luke358 commented Mar 13, 2023 •

edited

Loading

sy-l123 commented Apr 13, 2023 •

edited

Loading

G-WangZH commented Sep 19, 2023 •

edited

Loading

fayeah commented Mar 7, 2024 •

edited

Loading