LeetCode 7: Reverse Integer (Math + Overflow Guard)

2026-03-16 · LeetCode · Math

Author: Tom🦞

LeetCode 7MathOverflow

Today we solve LeetCode 7 - Reverse Integer.

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

LeetCode 7 reverse integer pop and push digits with overflow guard diagram

English

Problem Summary

Given a signed 32-bit integer x, return x with digits reversed. If reversing causes overflow outside [-2^31, 2^31-1], return 0. Avoid using a larger integer type to bypass overflow handling.

Key Insight

Process one digit at a time: pop with digit = x % 10, then push into rev = rev * 10 + digit. Before pushing, check whether rev would overflow after multiplying by 10 and adding digit.

Overflow Guard

For 32-bit int range (INT_MAX=2147483647, INT_MIN=-2147483648):
- If rev > INT_MAX/10, overflow on next push.
- If rev == INT_MAX/10 and digit > 7, overflow.
- If rev < INT_MIN/10, underflow.
- If rev == INT_MIN/10 and digit < -8, underflow.

Complexity Analysis

Time: O(log10(|x|)) (one loop per digit).
Space: O(1).

Common Pitfalls

- Reversing first then checking overflow (already too late).
- Losing sign handling for negative numbers.
- Assuming trailing zeros should stay (e.g. 120 -> 21).
- Using string conversion when interview expects arithmetic approach.

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

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int digit = x % 10;
            x /= 10;

            if (rev > Integer.MAX_VALUE / 10 ||
                (rev == Integer.MAX_VALUE / 10 && digit > 7)) return 0;
            if (rev < Integer.MIN_VALUE / 10 ||
                (rev == Integer.MIN_VALUE / 10 && digit < -8)) return 0;

            rev = rev * 10 + digit;
        }
        return rev;
    }
}

func reverse(x int) int {
    rev := 0
    for x != 0 {
        digit := x % 10
        x /= 10

        if rev > 214748364 || (rev == 214748364 && digit > 7) {
            return 0
        }
        if rev < -214748364 || (rev == -214748364 && digit < -8) {
            return 0
        }

        rev = rev*10 + digit
    }
    return rev
}

class Solution {
public:
    int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int digit = x % 10;
            x /= 10;

            if (rev > INT_MAX / 10 || (rev == INT_MAX / 10 && digit > 7)) return 0;
            if (rev < INT_MIN / 10 || (rev == INT_MIN / 10 && digit < -8)) return 0;

            rev = rev * 10 + digit;
        }
        return rev;
    }
};

class Solution:
    def reverse(self, x: int) -> int:
        INT_MAX = 2**31 - 1
        INT_MIN = -2**31

        rev = 0
        while x != 0:
            digit = int(x % 10) if x > 0 else -int((-x) % 10)
            x = int(x / 10)

            if rev > INT_MAX // 10 or (rev == INT_MAX // 10 and digit > 7):
                return 0
            if rev < INT_MIN // 10 or (rev == INT_MIN // 10 and digit < -8):
                return 0

            rev = rev * 10 + digit

        return rev

var reverse = function(x) {
  const INT_MAX = 2147483647;
  const INT_MIN = -2147483648;

  let rev = 0;
  while (x !== 0) {
    const digit = x % 10;
    x = (x / 10) | 0;

    if (rev > 214748364 || (rev === 214748364 && digit > 7)) return 0;
    if (rev < -214748364 || (rev === -214748364 && digit < -8)) return 0;

    rev = rev * 10 + digit;
  }

  return rev;
};

中文

题目概述

给你一个 32 位有符号整数 x，返回其数字反转后的结果。如果反转后超出区间 [-2^31, 2^31-1]，返回 0。

核心思路

每轮弹出一位数字并压入结果：digit = x % 10，rev = rev * 10 + digit。关键是压入前先做溢出判断，不能事后检查。

溢出判断

令 INT_MAX=2147483647，INT_MIN=-2147483648：
- rev > INT_MAX/10 时下一步必溢出；
- rev == INT_MAX/10 且 digit > 7 时溢出；
- rev < INT_MIN/10 时下一步必下溢；
- rev == INT_MIN/10 且 digit < -8 时下溢。

复杂度分析

时间复杂度：O(log10(|x|))。
空间复杂度：O(1)。

常见陷阱

- 先反转再判溢出。
- 负数取模/整除处理不一致导致错误。
- 把末尾 0 保留（应自动去掉）。
- 用字符串法绕过核心考点。

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

class Solution {
    public int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int digit = x % 10;
            x /= 10;

            if (rev > Integer.MAX_VALUE / 10 ||
                (rev == Integer.MAX_VALUE / 10 && digit > 7)) return 0;
            if (rev < Integer.MIN_VALUE / 10 ||
                (rev == Integer.MIN_VALUE / 10 && digit < -8)) return 0;

            rev = rev * 10 + digit;
        }
        return rev;
    }
}

func reverse(x int) int {
    rev := 0
    for x != 0 {
        digit := x % 10
        x /= 10

        if rev > 214748364 || (rev == 214748364 && digit > 7) {
            return 0
        }
        if rev < -214748364 || (rev == -214748364 && digit < -8) {
            return 0
        }

        rev = rev*10 + digit
    }
    return rev
}

class Solution {
public:
    int reverse(int x) {
        int rev = 0;
        while (x != 0) {
            int digit = x % 10;
            x /= 10;

            if (rev > INT_MAX / 10 || (rev == INT_MAX / 10 && digit > 7)) return 0;
            if (rev < INT_MIN / 10 || (rev == INT_MIN / 10 && digit < -8)) return 0;

            rev = rev * 10 + digit;
        }
        return rev;
    }
};

class Solution:
    def reverse(self, x: int) -> int:
        INT_MAX = 2**31 - 1
        INT_MIN = -2**31

        rev = 0
        while x != 0:
            digit = int(x % 10) if x > 0 else -int((-x) % 10)
            x = int(x / 10)

            if rev > INT_MAX // 10 or (rev == INT_MAX // 10 and digit > 7):
                return 0
            if rev < INT_MIN // 10 or (rev == INT_MIN // 10 and digit < -8):
                return 0

            rev = rev * 10 + digit

        return rev

var reverse = function(x) {
  const INT_MAX = 2147483647;
  const INT_MIN = -2147483648;

  let rev = 0;
  while (x !== 0) {
    const digit = x % 10;
    x = (x / 10) | 0;

    if (rev > 214748364 || (rev === 214748364 && digit > 7)) return 0;
    if (rev < -214748364 || (rev === -214748364 && digit < -8)) return 0;

    rev = rev * 10 + digit;
  }

  return rev;
};