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
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
一刷
题解：如果不停地拆分并求digit square sum， 可能陷入一种循环。如果最终收敛在1，那么表示是happy的，如果是不停地重复几个数字，那么是unhappy的。这是终止条件。于是这个可以借鉴linkedlist是否有环的快慢指针来做。
public class Solution {
private int digitSquareSum(int n){
int sum = 0, tmp;
while(n>0){
tmp = n%10;
sum += tmp*tmp;
n /= 10;
}
return sum;
}
public boolean isHappy(int n) {
int slow, fast;
slow = fast = n;
do{
slow = digitSquareSum(slow);
fast = digitSquareSum(fast);
fast = digitSquareSum(fast);
}while(slow!=fast);
if(slow == 1) return true;
else return false;
}
}
