1134. Armstrong Number

Description

Given an integer n, return true if and only if it is an Armstrong number.

The k-digit number n is an Armstrong number if and only if the kth power of each digit sums to n.

 

Example 1:

Input: n = 153
Output: true
Explanation: 153 is a 3-digit number, and 153 = 13 + 53 + 33.

Example 2:

Input: n = 123
Output: false
Explanation: 123 is a 3-digit number, and 123 != 13 + 23 + 33 = 36.

 

Constraints:

  • 1 <= n <= 108

Solutions

Solution 1: Simulation

We can first calculate the number of digits $k$, then calculate the sum $s$ of the $k$th power of each digit, and finally check whether $s$ equals $n$.

The time complexity is $O(\log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the given number.

Python Code
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class Solution:
    def isArmstrong(self, n: int) -> bool:
        k = len(str(n))
        s, x = 0, n
        while x:
            s += (x % 10) ** k
            x //= 10
        return s == n

Java Code
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class Solution {
    public boolean isArmstrong(int n) {
        int k = (n + "").length();
        int s = 0;
        for (int x = n; x > 0; x /= 10) {
            s += Math.pow(x % 10, k);
        }
        return s == n;
    }
}

C++ Code
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class Solution {
public:
    bool isArmstrong(int n) {
        int k = to_string(n).size();
        int s = 0;
        for (int x = n; x; x /= 10) {
            s += pow(x % 10, k);
        }
        return s == n;
    }
};

Go Code
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func isArmstrong(n int) bool {
	k := 0
	for x := n; x > 0; x /= 10 {
		k++
	}
	s := 0
	for x := n; x > 0; x /= 10 {
		s += int(math.Pow(float64(x%10), float64(k)))
	}
	return s == n
}

TypeScript Code
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function isArmstrong(n: number): boolean {
    const k = String(n).length;
    let s = 0;
    for (let x = n; x; x = Math.floor(x / 10)) {
        s += Math.pow(x % 10, k);
    }
    return s == n;
}

JavaScript Code
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/**
 * @param {number} n
 * @return {boolean}
 */
var isArmstrong = function (n) {
    const k = String(n).length;
    let s = 0;
    for (let x = n; x; x = Math.floor(x / 10)) {
        s += Math.pow(x % 10, k);
    }
    return s == n;
};