LeetCode 202. Happy Number

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

思路：

将一个数的各个位上的数平方相加后得到新数，看最后能不能变成1，如果能返回true,否则返回false

如果不能变成1，最后其实是一个循环，取了一个循环中的数89来作为是否一直在循环里跑的条件。

代码：

class Solution {
public:
bool isHappy(int n) {
int result = n;
vector<int> a;
while (result != 89){//89在循环中，用来作为是否能变成1的条件之一
while (result){//将result的各个位保存到vector<int> a中
a.push_back(result % 10);
result /= 10;
}

for (vector<int>::iterator ix = a.begin(); ix != a.end(); ++ix){//计算新的result值
result += (*ix)*(*ix);
}
a.clear();//a一定要清零

if (result == 1){//result如果为1了返回true
return true;
}
}
return false;//如果进入了计算数的循环，也就是有89出现，返回false
}
};