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

Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.

写一个算法判断一个数是不是快乐数。快乐数：一个正整数，如果对其各个位上的数字分别平方求和得到一个新的数，再进行同样的操作，如果结果变成了1，则说明是快乐数。

解法：循环求平方和，即求出当前数组的平方和后，再以此平方和作为新的数继续求平方和，而循环终止条件是：得到的平方和为1，或得到的平方和在之前的循环中出现过，那以后会一直循环，不可能达到1。判断平方和是否为1很简单，每次检查就好了；而判断平方和是否出现过，则只需要维持一个Set，每次循环检查当前平方和是否在Set中，在则终止循环，不在则将此平方和放到Set中。

Java：

public class Solution {
public boolean isHappy(int n) {
Set<Integer> got = new HashSet<>();
while (n != 1 && !got.contains(n)) {
got.add(n);
int sum = 0;
while (n != 0) {
sum += Math.pow(n % 10, 2);
n /= 10;
}
n = sum;
}
return n == 1;
}
}

Python:

class Solution:
# @param {integer} n
# @return {boolean}
def isHappy(self, n):
lookup = {}
while n != 1 and n not in lookup:
lookup
 = True
n = self.nextNumber(n)
return n == 1

def nextNumber(self, n):
new = 0
for char in str(n):
new += int(char)**2
return new　　

Python:

class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
got = set()
while n != 1 and n not in got:
got.add(n)
sum = 0
while n:
sum += (n % 10)**2
n //= 10
n = sum

return n == 1

C++:

class Solution {
public:
bool isHappy(int n) {
set<int> got;
while (n != 1 && got.find(n) == got.end()) {
got.insert(n);
int sum = 0;
while (n) {
sum += pow(n % 10, 2);
n /= 10;
}
n = sum;
}
return n == 1;
}
};

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