Question
Write an algorithm to determine if a number is happy.
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example
10 is a happy number
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
Thinking
- All negative and zero are not happy number.
- One is a happy number.
- Using % and / to get each number by postion.
- Check whether there is a cycle of calculation. (Result appears before)
Review
If any result is 4, then the number is not a happy number.
Happy number wiki
Other solutions
Solution
Java (Review, passed on leetcode)
public class Solution {
public boolean isHappy(int n) {
if (n < 0) {
return false;
}
Set<Integer> results = new HashSet<Integer>();
results.add(n);
while (n != 1) {
int sum = 0;
while (n != 0) {
sum += (n % 10) * (n % 10);
n = n / 10;
}
n = sum;
if (results.contains(n)) {
return false;
} else {
results.add(n);
}
}
return true;
}
}
Java (Review, check sum == 4, passed on leetcode)
public class Solution {
public boolean isHappy(int n) {
if (n < 0) {
return false;
}
while (n != 1) {
int sum = 0;
while (n != 0) {
sum += (n % 10) * (n % 10);
n = n / 10;
}
n = sum;
if (n == 4) {
return false;
}
}
return true;
}
}
java
public class Solution {
/**
* @param n an integer
* @return true if this is a happy number or false
*/
public boolean isHappy(int n) {
// Write your code here
if (n <= 0) {
return false;
}
if (n == 1) {
return true;
}
Map<Integer, Integer> sumMap = new HashMap<Integer, Integer>();
sumMap.put(n, 1);
while (true) {
int sum = 0;
if (n < 10) {
sum = n * n;
} else {
int n1 = n % 10;
int n2 = n / 10;
while (n2 > 0) {
sum += n1 * n1;
n1 = n2 % 10;
n2 = n2 / 10;
}
sum += n1 * n1;
}
if (sum == 1) {
return true;
}
if (sumMap.get(sum) != null) {
return false;
}
sumMap.put(sum, 1);
n = sum;
}
}
}