LeetCode 2119: A Number After a Double Reversal (Trailing Zero Invariant)

Today we solve LeetCode 2119 - A Number After a Double Reversal.

Source: https://leetcode.com/problems/a-number-after-a-double-reversal/

English

Problem Summary

Given a non-negative integer num, reverse its digits twice. Return true if the final value equals the original num; otherwise return false.

Key Insight

The first reversal removes leading zeros in the reversed number, which correspond to trailing zeros of the original number. So any positive number ending with 0 will lose digits and cannot recover after the second reversal. The only exception is 0 itself.

Algorithm

- If num == 0, return true.
- Otherwise, return num % 10 != 0.

Complexity Analysis

Time: O(1).
Space: O(1).

Common Pitfalls

- Simulating full reversal with loops is unnecessary.
- Forgetting the special case num = 0 (it should be true).
- Mistakenly thinking all even numbers fail (only numbers ending in zero fail, except 0).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public boolean isSameAfterReversals(int num) {
        return num == 0 || num % 10 != 0;
    }
}

func isSameAfterReversals(num int) bool {
    return num == 0 || num%10 != 0
}

class Solution {
public:
    bool isSameAfterReversals(int num) {
        return num == 0 || num % 10 != 0;
    }
};

class Solution:
    def isSameAfterReversals(self, num: int) -> bool:
        return num == 0 or num % 10 != 0

var isSameAfterReversals = function(num) {
  return num === 0 || num % 10 !== 0;
};

中文

题目概述

给你一个非负整数 num，把它的十进制数字翻转两次。若最终结果与原数相同返回 true，否则返回 false。

核心思路

第一次翻转会去掉翻转后数字的前导零，而这些前导零正好来自原数的尾随零。因此，所有结尾是 0 的正整数在第一次翻转后会“丢位”，第二次翻转也无法恢复。唯一例外是 0 本身。

算法步骤

- 若 num == 0，返回 true。
- 否则判断 num % 10 != 0。

复杂度分析

时间复杂度：O(1)。
空间复杂度：O(1)。

常见陷阱

- 没必要真的写循环做两次翻转。
- 忘记处理 num = 0 这个特例（应返回 true）。
- 误以为所有偶数都不行；实际上只和“末位是否为 0”有关。

多语言参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public boolean isSameAfterReversals(int num) {
        return num == 0 || num % 10 != 0;
    }
}

func isSameAfterReversals(num int) bool {
    return num == 0 || num%10 != 0
}

class Solution {
public:
    bool isSameAfterReversals(int num) {
        return num == 0 || num % 10 != 0;
    }
};

class Solution:
    def isSameAfterReversals(self, num: int) -> bool:
        return num == 0 or num % 10 != 0

var isSameAfterReversals = function(num) {
  return num === 0 || num % 10 !== 0;
};