我想这个解决办法会适合你

const a = [1, 2, 3, 2];
const b = ['this', 'is', 'my', 'good example'];

const c = a.flatMap((count, i) => Array(count).fill(b[i]));

console.log(c);

你可以在另一个循环中做一个循环来实现

const a = [1, 2, 3, 2];
const b = ['this', 'is', 'my', 'good example'];

let array = [];
b.forEach((x, i) => {
  for (let j = 0; j < a[i]; j++) {
    array.push(x);
  }
})

console.log(array)

const a = [1, 2, 3],
      b = ['a','b','c']

console.log(
  b.reduce((acc, item, idx) => acc.concat(Array(a[idx]).fill(item)), [])
)

// or the opposite

console.log(
  a.reduce((acc, item, idx) => acc.concat(Array(item).fill(b[idx])), [])
)

我同意flatMap是完成这项任务的理想选择，@lvjonok对此做出了回答

您可以使用Array#reduce方法，如下所示：

const
      a = [ 1, 2, 3, 2],
      b = ['this', 'is', 'my', 'good example'],
      
      c = a.reduce(
          (acc,cur,i) => [ ...acc, ...Array(cur).fill( b[i] ) ], []
      );
      
console.log( c );