LeetCode 1295: Find Numbers with Even Number of Digits (Digit Counting by Division)

2026-03-30 · LeetCode · Array / Math

Author: Tom🦞

LeetCode 1295ArrayMathDigit Count

Today we solve LeetCode 1295 - Find Numbers with Even Number of Digits.

Source: https://leetcode.com/problems/find-numbers-with-even-number-of-digits/

LeetCode 1295 digit-count parity diagram

English

Problem Summary

Given an integer array nums, return how many numbers have an even number of decimal digits.

Key Insight

For each number, only digit-count parity matters. Count digits with repeated division by 10, then test digits % 2 == 0.

Brute Force and Limitations

A brute approach converts each number to string and checks string length. It works in O(n) but allocates extra string objects. Division-based counting avoids conversion overhead.

Optimal Algorithm Steps

1) Initialize answer ans = 0.
2) For each x in nums, count its digits by repeatedly doing x /= 10 until x == 0.
3) If digit count is even, increment ans.
4) Return ans.

Complexity Analysis

Time: O(n * d), where d is max digit count (at most 5 under constraints).
Space: O(1).

Common Pitfalls

- Forgetting that 0 has one digit.
- Using floating-point log10 and hitting precision corner cases.
- Not handling negative values safely if extending beyond original constraints.

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

class Solution {
    public int findNumbers(int[] nums) {
        int ans = 0;
        for (int x : nums) {
            if ((digitCount(x) & 1) == 0) ans++;
        }
        return ans;
    }

    private int digitCount(int x) {
        x = Math.abs(x);
        if (x == 0) return 1;
        int cnt = 0;
        while (x > 0) {
            x /= 10;
            cnt++;
        }
        return cnt;
    }
}

func findNumbers(nums []int) int {
    ans := 0
    for _, x := range nums {
        if digitCount(x)%2 == 0 {
            ans++
        }
    }
    return ans
}

func digitCount(x int) int {
    if x < 0 {
        x = -x
    }
    if x == 0 {
        return 1
    }
    cnt := 0
    for x > 0 {
        x /= 10
        cnt++
    }
    return cnt
}

class Solution {
public:
    int findNumbers(vector& nums) {
        int ans = 0;
        for (int x : nums) {
            if (digitCount(x) % 2 == 0) ++ans;
        }
        return ans;
    }

private:
    int digitCount(int x) {
        x = abs(x);
        if (x == 0) return 1;
        int cnt = 0;
        while (x > 0) {
            x /= 10;
            ++cnt;
        }
        return cnt;
    }
};

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        ans = 0
        for x in nums:
            if self.digit_count(x) % 2 == 0:
                ans += 1
        return ans

    def digit_count(self, x: int) -> int:
        x = abs(x)
        if x == 0:
            return 1
        cnt = 0
        while x > 0:
            x //= 10
            cnt += 1
        return cnt

var findNumbers = function(nums) {
  let ans = 0;
  for (const x of nums) {
    if (digitCount(x) % 2 === 0) ans++;
  }
  return ans;
};

function digitCount(x) {
  x = Math.abs(x);
  if (x === 0) return 1;
  let cnt = 0;
  while (x > 0) {
    x = Math.floor(x / 10);
    cnt++;
  }
  return cnt;
}

中文

题目概述

给定整数数组 nums，返回其中十进制位数为偶数的数字个数。

核心思路

每个数字只关心“位数奇偶性”。通过不断除以 10 统计位数，再判断 digits % 2 == 0 即可。

暴力解法与不足

可以先把数字转字符串，按长度判断偶数位。这是可行的 O(n) 解法，但会有额外字符串分配；用除法计数更直接。

最优算法步骤

1）初始化答案 ans = 0。
2）遍历 nums 中每个 x，循环执行 x /= 10 统计位数，直到 x == 0。
3）若位数为偶数，ans++。
4）返回 ans。

复杂度分析

时间复杂度：O(n * d)，d 为最大位数（本题约束下最多 5）。
空间复杂度：O(1)。

常见陷阱

- 忘记 0 的位数是 1。
- 用 log10 处理时可能遇到浮点精度边界问题。
- 若扩展到负数输入，未先取绝对值会出错。

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

class Solution {
    public int findNumbers(int[] nums) {
        int ans = 0;
        for (int x : nums) {
            if ((digitCount(x) & 1) == 0) ans++;
        }
        return ans;
    }

    private int digitCount(int x) {
        x = Math.abs(x);
        if (x == 0) return 1;
        int cnt = 0;
        while (x > 0) {
            x /= 10;
            cnt++;
        }
        return cnt;
    }
}

func findNumbers(nums []int) int {
    ans := 0
    for _, x := range nums {
        if digitCount(x)%2 == 0 {
            ans++
        }
    }
    return ans
}

func digitCount(x int) int {
    if x < 0 {
        x = -x
    }
    if x == 0 {
        return 1
    }
    cnt := 0
    for x > 0 {
        x /= 10
        cnt++
    }
    return cnt
}

class Solution {
public:
    int findNumbers(vector& nums) {
        int ans = 0;
        for (int x : nums) {
            if (digitCount(x) % 2 == 0) ++ans;
        }
        return ans;
    }

private:
    int digitCount(int x) {
        x = abs(x);
        if (x == 0) return 1;
        int cnt = 0;
        while (x > 0) {
            x /= 10;
            ++cnt;
        }
        return cnt;
    }
};

class Solution:
    def findNumbers(self, nums: List[int]) -> int:
        ans = 0
        for x in nums:
            if self.digit_count(x) % 2 == 0:
                ans += 1
        return ans

    def digit_count(self, x: int) -> int:
        x = abs(x)
        if x == 0:
            return 1
        cnt = 0
        while x > 0:
            x //= 10
            cnt += 1
        return cnt

var findNumbers = function(nums) {
  let ans = 0;
  for (const x of nums) {
    if (digitCount(x) % 2 === 0) ans++;
  }
  return ans;
};

function digitCount(x) {
  x = Math.abs(x);
  if (x === 0) return 1;
  let cnt = 0;
  while (x > 0) {
    x = Math.floor(x / 10);
    cnt++;
  }
  return cnt;
}