题目
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: 19 is a happy number
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.
分析
此题目，只要理解了快乐数的判定条件，便很简单了。
快乐数（happy number）有以下的特性：在给定的进位制下，该数字所有数位(digits)的平方和，得到的新数再次求所有数位的平方和，如此重复进行，最终结果必为1。
以十进位为例：
2 8 → 2^2+8^2=68 → 6^2+8^2=100 → 1^2+0^2+0^2=1
3 2 → 3^2+2^2=13 → 1^2+3^2=10 → 1^2+0^2=1
3 7 → 3^2+7^2=58 → 5^2+8^2=89 → 8^2+9^2=145 → 1^2+4^2+5^2=42 → 4^2+2^2=20 → 2^2+0^2=4 → 4^2=16 → 1^2+6^2=37……
因此28和32是快乐数，而在37的计算过程中，37重覆出现，继续计算的结果只会是上述数字的循环，不会出现1，因此37不是快乐数。
不是快乐数的数称为不快乐数（unhappy number），所有不快乐数的数位平方和计算，最後都会进入 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 的循环中。
因此，只要循环计算过程中，结果为该数最初值或4，则必然为不快乐数。
AC代码
class Solution {
public:
bool isHappy(int n) {
if (n <= 0)
return false;
if (n == 1)
return true;
int sum = n;
while (sum != 1)
{
int tmp = sum;
sum = 0;
//求各个位平方和
while (tmp != 0)
{
sum += pow(tmp % 10, 2);
tmp /= 10;
}//while
if (sum == n || sum == 4)
return false;
}//while
if (sum == 1)
return true;
else
return false;
}
};