1134

1134. Armstrong Number (Easy)

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

Companies:
Amazon

Related Topics:
Math

Solution 1.

// OJ: https://leetcode.com/problems/armstrong-number/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/armstrong-number/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool isArmstrong(int n) {
        long sum = 0, cnt = log10(n) + 1, target = n;
        while (n) {
            sum += pow(n % 10, cnt);
            n /= 10;
        }
        return sum == target;
    }
};