0007. Reverse Integer (E)

Reverse Integer (E)

题目

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: \([-2^{31}, 2^{31}-1]\). For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

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题意

给定一个32位有符号整数，将其各位倒置后输出，如果倒置后超出int范围溢出则输出0。

思路

问题的关键在于对溢出数的处理。溢出后再与最值比较不可取，因此通过与最值/10比较来判断最后会不会溢出。int范围为[-2147483648, 2147483647]。

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代码实现

Java

class Solution {
    public int reverse(int x) {
        int res = 0;
        while (x != 0) {
            int n = x % 10;
            if ((res > Integer.MAX_VALUE / 10)
                    || (res == Integer.MAX_VALUE / 10 && n > 7)) {
                return 0;
            } else if ((res < Integer.MIN_VALUE / 10)
                    || (res == Integer.MIN_VALUE / 10 && n < -8)) {
                return 0;
            } else {
                res = 10 * res + n;
            }
            x /= 10;
        }
        return res;
    }
}

JavaScript

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function (x) {
  let minVal = -Math.pow(2, 31), maxVal = Math.pow(2, 31) - 1
  let res = 0
  while (x !== 0) {
    let num = x % 10
    if ((res > maxVal / 10 || res === Math.trunc(maxVal / 10) && num > 7) ||
        (res < minVal / 10 || res === Math.trunc(minVal / 10) && num < -8)) {
      return 0
    }
    res = res * 10 + num
    x = Math.trunc(x / 10)
  }
  return res
}