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:
Solutions
Solution 1: Simulation
We can first calculate the number of digits k, then calculate the sum s of the kth 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.
PythonJavaC++GoTypeScriptJavaScript
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
/**
* @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;
};(code-box)
