Factor Combinations

Description

null

Solution(javascript)

// /** 1: 手动去重的版本
//  * @param {number} n
//  * @return {number[][]}
//  */
// const getFactors = function (n) {
//   const result = {}
//   const aux = (remain, current = []) => {
//     if (remain === 1) {
//       if (current.length > 1) {
//         current.sort((a, b) => a - b)
//         const key = current.join('')
//         result[key] = current
//       }
//       return
//     }
//     for (let i = 2; i <= remain; i++) {
//       if (remain % i === 0) {
//         aux(remain / i, [...current, i])
//       }
//     }
//   }
//   aux(n)
//   return Object.values(result)
// }


/** 2: 控制边界去重
 * @param {number} n
 * @return {number[][]}
 */
const getFactors = function (n) {
  const result = []
  const aux = (remain, start = 2, current = []) => {
    if (remain === 1) {
      if (current.length > 1) {
        result.push([...current])
      }
      return
    }
    for (let i = start; i <= remain; i++) {
      if (remain % i === 0) {
        current.push(i)
        aux(remain / i, i, current)
        current.pop()
      }
    }
  }
  aux(n)
  return result
}