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
思路：
这题目挺有意思的，肯定不能真的通过无限循环来判断。我观察了一下，估计不happy的数字，运算一圈后会出现前面已经出现过的数字，即在一组数字里绕圈圈。故记录一下出现过的数字，判断是否重复。
如果重复了就不happy，得到1了就happy。
bool isHappy(int n) {
unordered_set<int> record;
while()
{
int sum = ;
while(n > )
{
sum += (n % ) * (n % );
n /= ;
}
if(sum == )
return true;
if(record.find(sum) == record.end()) //当前数字没有出现过
{
record.insert(sum);
n = sum;
}
else
return false;
}
}
还有用O(1)空间的，就像链表找有没有圈一样，用快慢指针的思路。
public class Solution {
public boolean isHappy(int n) {
int x = n;
int y = n;
while(x>1){
x = cal(x) ;
if(x==1) return true ;
y = cal(cal(y));
if(y==1) return true ;
if(x==y) return false;
}
return true ;
}
public int cal(int n){
int x = n;
int s = 0;
while(x>0){
s = s+(x%10)*(x%10);
x = x/10;
}
return s ;
}
}
还有用数学的，所有不happy的数字，都会得到4.（？？why）
bool isHappy(int n) {
if (n <= ) return false;
int magic = ;
while () {
if (n == ) return true;
if (n == magic) return false;
int t = ;
while (n) {
t += (n % ) * (n % );
n /= ;
}
n = t;
}
}