LeetCode 202. Happy Number (Easy)

题目描述：

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

题目大意：判断一个数字是否“开心数”：每一位的平方的和等于1就是开心数，否则用该和继续运算。

思路：一直循环运算就可以了，但是怎么让循环终止呢（终止条件是和为1或者找不到这个数），用一个set把之前的结果存起来，如果发现结果和之前的结果相同就说明进入了“循环“的状态。

然而既然如此，为什么不用寻环算法呢…..直接快慢指针相遇就可以知道有没有环了。

c++代码：

class Solution {
public:
bool isHappy(int n) {
set<int> s;
while (1)
{
n = getNums(n);
if (n == 1)
return true;
else
{
if (s.count(n) == 0)
s.insert(n);
else
return false;
}
}
}
int getNums(int n)
{
int ret = 0;
while (n != 0)
{
ret += (n % 10) * (n % 10);
n /= 10;
}
return ret;
}
};