Question

Formatted question description: https://leetcode.ca/all/1134.html

 1134. Armstrong Number

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

 Given a positive integer N, return true if and only if it is an Armstrong number.

 Example 1:

 Input: 153
 Output: true
 Explanation:
 153 is a 3-digit number, and 153 = 1^3 + 5^3 + 3^3.

 Example 2:

 Input: 123
 Output: false
 Explanation:
 123 is a 3-digit number, and 123 != 1^3 + 2^3 + 3^3 = 36.

 Note:

 1 <= N <= 10^8

Algorithm

Just follow problem description.

Code

public class Armstrong_Number {

    class Solution {
        public boolean isArmstrong(int N) {
            int sum = 0;
            int tmp = N;
            int length = Integer.toString(N).length();
            while (tmp > 0){
                sum += power(tmp % 10, length);
                tmp /= 10;
            }
            return sum == N;
        }

        private int power(int a, int b){
            int res = 1;
            for (int i = 0; i < b; i++){
                res *= a;
            }
            return res;
        }
    }

}

// OJ: https://leetcode.com/problems/armstrong-number/
// Time: O(N)
// Space: O(lgN)
class Solution {
public:
    bool isArmstrong(int n) {
        long sum = 0, target = n;
        vector<int> digits;
        while (n) {
            digits.push_back(n % 10);
            n /= 10;
        }
        for (int d : digits) sum += pow(d, digits.size());
        return sum == target;
    }
};

class Solution:
    def isArmstrong(self, n: int) -> bool:
        k = len(str(n))
        s, t = 0, n
        while t:
            t, v = divmod(t, 10)
            s += pow(v, k)
        return n == s

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
}

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;
}

/**
 * @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;
};

1134 - Armstrong Number

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