有规律的合并两个数组

rubinTime · 发表于 2022-2-24 10:52:21

//有规律的合并两个数组
// ['A1','A2','A3','B1','B2','C1','C2','C3','D1','D2']-数组1
// ['A','B','C','D']-数组2
// 合并为 ['A1','A2','A3','A','B1','B2',"B",'C1','C2','C3',"C",'D1','D2',"D"]
const easyCloneObj=obj=>JSON.parse(JSON.stringify(obj)) 

// 1.遍历数组2，使用数组2的元素作为正则，匹配数组1
function combine1(arr1,arr2){
     let currentIndex=0
     let result=easyCloneObj(arr1)
    for(let i=0;i<arr2.length;i++){
       const reg=new RegExp(arr2[i])
       while(currentIndex++<result.length){
            if(!reg.test(result[currentIndex])){
              result.splice(currentIndex,0,arr2[i])
              break;
            }
       }
    }
    // console.log(arr1)
    return result
}
// const arr3=combine1(['A1','A2','A3','B1','B2','C1','C2','C3','D1','D2'],['A','B','C','D'])
// console.log(arr3)

//遍历数组1，使用数组1首字符去匹配数组二的元素
function combine2(arr1,arr2){
   let result=[]
   let j=0
   for(let i=0;i<arr1.length;i++){
     const char=arr1[i][0]
     if(char==arr2[j]){
       result.push(arr1[i])
     }else{
       result.push(arr2[j])
       result.push(arr1[i])
       j++
     }
     //当最后一轮循环中不会再有别的字符，进入不了else分支
     //导致arr2的最后一个元素没有push进来需要特殊处理
     if(i==arr1.length-1){
       result.push(arr2[j])
     }
   }
   return result
}
const arr4=combine2(['A1','A2','A3','B1','B2','C1','C2','C3','D1','D2'],['A','B','C','D'])
console.log(arr4)