题解 | #数组分组#

//参考网友ld1230的解法
//思想：将能整除3或者5的各自分为一组，记为sum1和sum2，剩余的保存在数组nums里
//现有两组，剩余nums里的数要么在sum1里要么在sum2里，利用递归依次放在sum1和sum2中
//最终nums里的数字全部放进去，若sum1 == sum2，则返回true，否则，返回false
let str;
while(str = readline()){
  let num = readline();
  let arr = num.split(' ');
  let three = arr.filter((v)=>{
    return v%3==0 && v%5!=0;
  })

  let five = arr.filter((v)=>{
    return v%5==0;
  })

  let other = arr.filter((v)=>{
    return v%5!=0 &&v%3!=0;
  })
  //所有三的倍数的数字的和
  let tsum = three.reduce((pre,cur)=>{
    return pre+cur*1;
  },0)
  //所有五的倍数的数字的和
  let fsum = five.reduce((pre,cur)=>{
    return pre+cur*1;
  },0)

  console.log(isExists(tsum,fsum,other,0))
}

function isExists(sum1,sum2,nums,index){
    if(index == nums.length && sum1 != sum2) return false;
    if(index == nums.length && sum1 == sum2) return true;
    if(index < nums.length) return isExists(sum1+parseInt(nums[index]), sum2, nums, index+1) || isExists(sum1, sum2+parseInt(nums[index]), nums, index+1);
    return false;
}