陈恒承

7. 整数反转

https://leetcode-cn.com/problems/reverse-integer/

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function (x) {
  let islowerZero = x < 0;

  if (islowerZero) {
    x = Math.abs(x);
  }

  let len = x.toString().length - 1;

  let reverseNumber = 0;

  while (x) {
    let rest = x % 10;
    reverseNumber += rest * Math.pow(10, len);

    x = Math.floor(x / 10);
    len--;
  }

  if (reverseNumber > Math.pow(2, 31) - 1) return 0;

  return islowerZero ? -reverseNumber : reverseNumber;
};

上面的方式不是不可以，但我没有发现一个规律,那就是 res=res*10+rest(余数)

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function (x) {
  let islowerZero = x < 0;

  if (islowerZero) {
    x = Math.abs(x);
  }
  let reverseNumber = 0;
  while (x) {
    let rest = x % 10;
    reverseNumber += reverseNumber * 10 + rest;
    x = Math.floor(x / 10);
  }

  if (reverseNumber > Math.pow(2, 31) - 1) return 0;

  return islowerZero ? -reverseNumber : reverseNumber;
};