style: format code (#4212)

close #4204
This commit is contained in:
acbin
2023-06-09 18:52:05 +08:00
committed by GitHub
parent ad03086f54
commit 00282efd8b
521 changed files with 5233 additions and 7309 deletions

View File

@@ -1,7 +1,9 @@
/** Author : Suraj Kumar Modi
/**
* Author : Suraj Kumar Modi
* https://github.com/skmodi649
*/
/** You are given a number n. You need to find the digital root of n.
/**
* You are given a number n. You need to find the digital root of n.
* DigitalRoot of a number is the recursive sum of its digits until we get a single digit number.
*
* Test Case 1:
@@ -19,7 +21,8 @@
* sum of digit of 45 is 9 which is a single
* digit number.
*/
/** Algorithm :
/**
* Algorithm :
* Step 1 : Define a method digitalRoot(int n)
* Step 2 : Define another method single(int n)
* Step 3 : digitalRoot(int n) method takes output of single(int n) as input
@@ -32,14 +35,16 @@
* return n;
* else
* return (n%10) + (n/10)
* Step 5 : In main method simply take n as input and then call digitalRoot(int n) function and print the result
* Step 5 : In main method simply take n as input and then call digitalRoot(int n) function and
* print the result
*/
package com.thealgorithms.maths;
class DigitalRoot {
public static int digitalRoot(int n) {
if (single(n) <= 9) { // If n is already single digit than simply call single method and return the value
if (single(n) <= 9) { // If n is already single digit than simply call single method and
// return the value
return single(n);
} else {
return digitalRoot(single(n));
@@ -55,7 +60,6 @@ class DigitalRoot {
}
} // n / 10 is the number obtainded after removing the digit one by one
// Sum of digits is stored in the Stack memory and then finally returned
}
/**
* Time Complexity : O((Number of Digits)^2) Auxiliary Space Complexity :