1-1000回文数

[Sunny-117]

No description provided.

[mengqiuleo]

function getReverseNumber(num){
  const res = [];
  for(let i=10; i<=num; i++){
    let a = i.toString();
    let b = a.split('').reverse().join('');
    if(a == b){
      res.push(b);
    }
  }
  return res;
}

console.log(getReverseNumber(1000))

[bearki99]

function isReverse(num) {
  let newNum = num + "";
  let l = 0,
    r = newNum.length - 1;
  while (l < r) {
    if (newNum[l] !== newNum[r]) return false;
    l++;
    r--;
  }
  return true;
}
const res = [];
for (let i = 1; i <= 1000; i++) {
  if (isReverse(i)) res.push(i);
}
console.log(res);

[HangHZhang]

function palindromes1(num) {
    return +((''+num).split('').reverse().join('')) === num
}

[HangHZhang]

function palindromes2(num) {
    const numStr = '' + num
    let left = 0;
    let right = numStr.length - 1
    while ( left <= right) {
        if (numStr[left++] === numStr[right--]) {
            continue
        } else {
            return false
        }
    }
    return true
}