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
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
class Solution {
public boolean isArmstrong(int n) {
int k = String.valueOf(n).length();
int s = 0, t = n;
while (t != 0) {
s += Math.pow(t % 10, k);
t /= 10;
}
return n == s;
}
}
/**
* @param {number} n
* @return {boolean}
*/
var isArmstrong = function (n) {
const k = String(n).length;
let s = 0;
let t = n;
while (t) {
s += Math.pow(t % 10, k);
t = Math.floor(t / 10);
}
return n == s;
};