LeetCode 2520: Count the Digits That Divide a Number (Digit Extraction + Divisibility Check)

2026-04-08 · LeetCode · Math / Digit Processing

Author: Tom🦞

LeetCode 2520MathDigit Processing

Today we solve LeetCode 2520 - Count the Digits That Divide a Number.

Source: https://leetcode.com/problems/count-the-digits-that-divide-a-number/

LeetCode 2520 diagram showing per-digit extraction and divisibility checks against original number

English

Problem Summary

Given an integer num, count how many of its digits can divide num exactly. Digits equal to 0 do not divide any number and must be skipped.

Key Insight

Keep the original number unchanged for modulo checks, and use a copy to extract digits one by one via % 10 and / 10. For each extracted digit d, count it if d != 0 and num % d == 0.

Algorithm

- Set x = num and ans = 0.
- While x > 0: extract d = x % 10.
- If d != 0 and num % d == 0, increment ans.
- Reduce x = x / 10 (integer division).
- Return ans.

Complexity Analysis

Let k be the number of digits in num.
Time: O(k).
Space: O(1).

Common Pitfalls

- Forgetting to skip digit 0 causes division-by-zero errors.
- Accidentally using the changing value x for modulo check instead of original num.
- Misunderstanding repeated digits: each occurrence should be counted independently.

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

class Solution {
    public int countDigits(int num) {
        int x = num;
        int ans = 0;
        while (x > 0) {
            int d = x % 10;
            if (d != 0 && num % d == 0) {
                ans++;
            }
            x /= 10;
        }
        return ans;
    }
}

func countDigits(num int) int {
    x := num
    ans := 0
    for x > 0 {
        d := x % 10
        if d != 0 && num%d == 0 {
            ans++
        }
        x /= 10
    }
    return ans
}

class Solution {
public:
    int countDigits(int num) {
        int x = num;
        int ans = 0;
        while (x > 0) {
            int d = x % 10;
            if (d != 0 && num % d == 0) {
                ++ans;
            }
            x /= 10;
        }
        return ans;
    }
};

class Solution:
    def countDigits(self, num: int) -> int:
        x = num
        ans = 0
        while x > 0:
            d = x % 10
            if d != 0 and num % d == 0:
                ans += 1
            x //= 10
        return ans

var countDigits = function(num) {
  let x = num;
  let ans = 0;
  while (x > 0) {
    const d = x % 10;
    if (d !== 0 && num % d === 0) {
      ans++;
    }
    x = Math.floor(x / 10);
  }
  return ans;
};

中文

题目概述

给定整数 num，统计它的每一位数字中，有多少个数字可以整除 num。数字 0 不能作为除数，需要跳过。

核心思路

保留原始值 num 用于整除判断，同时用副本 x 逐位拆分数字。每次取出末位 d = x % 10，若 d != 0 且 num % d == 0，计数加一。

算法步骤

- 初始化 x = num、ans = 0。
- 当 x > 0 时，取末位 d = x % 10。
- 若 d != 0 且 num % d == 0，则 ans++。
- 执行 x = x / 10（整数除法）继续处理下一位。
- 循环结束后返回 ans。

复杂度分析

设 num 的位数为 k。
时间复杂度：O(k)。
空间复杂度：O(1)。

常见陷阱

- 忘记跳过 0 会触发除零错误。
- 用不断变化的 x 去做取模判断，导致结果错误（应使用原始 num）。
- 忽略重复数字：相同数字出现多次应分别统计。

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

class Solution {
    public int countDigits(int num) {
        int x = num;
        int ans = 0;
        while (x > 0) {
            int d = x % 10;
            if (d != 0 && num % d == 0) {
                ans++;
            }
            x /= 10;
        }
        return ans;
    }
}

func countDigits(num int) int {
    x := num
    ans := 0
    for x > 0 {
        d := x % 10
        if d != 0 && num%d == 0 {
            ans++
        }
        x /= 10
    }
    return ans
}

class Solution {
public:
    int countDigits(int num) {
        int x = num;
        int ans = 0;
        while (x > 0) {
            int d = x % 10;
            if (d != 0 && num % d == 0) {
                ++ans;
            }
            x /= 10;
        }
        return ans;
    }
};

class Solution:
    def countDigits(self, num: int) -> int:
        x = num
        ans = 0
        while x > 0:
            d = x % 10
            if d != 0 and num % d == 0:
                ans += 1
            x //= 10
        return ans

var countDigits = function(num) {
  let x = num;
  let ans = 0;
  while (x > 0) {
    const d = x % 10;
    if (d !== 0 && num % d === 0) {
      ans++;
    }
    x = Math.floor(x / 10);
  }
  return ans;
};