Leetcode 202. Happy Number

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: 

Input: 19
Output: true
Explanation: 
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

解析：快乐数，给一个数字n，将每个位置上的数字进行平方求和，循环下去，如果和为1则为快乐数，否则无限循环。首先看一下非快乐数的情况。

由上图可以发现，如果是非快乐数，则后面的数字一直在循环，所以可以设置一个HashSet，将出现的数字存放进去，如果出现重复的则跳出循环，如果为1，则为true。具体代码如下：

class Solution {
public:
    bool isHappy(int n) {
        unordered_set<int> sets;
        int sum = 0;
        while(n != 1)
        {
            sum = get_square_sum(n);
            if (sets.count(sum))
                break;
            sets.insert(sum);
            n = sum;
        }
        return n ==1;
    }
    int get_square_sum(int n)
    {
        int sum = 0;
        while(n != 0)
        {
            int single_num = n%10;
            sum += pow(single_num, 2);
            n /= 10;
        }
        return sum;
    }
    
};

解法2：

采用检测链表环的思路，这个题目中，数字一定存在部分重复，可以认为是存在环，然后采用链表检测环的思路，使用快慢指针来做。

class Solution {
public:
    bool isHappy(int n) {
        /**
        unordered_set<int> sets;
        int sum = 0;
        while(n != 1)
        {
            sum = get_square_sum(n);
            if (sets.count(sum))
                break;
            sets.insert(sum);
            n = sum;
        }
        return n ==1;
        **/
        int slow, fast;
        slow = fast = n;
        do
        {
            slow = get_square_sum(slow);
            fast = get_square_sum(fast);
            fast = get_square_sum(fast);
        }while(slow != fast);
        if(slow == 1)
            return true;
        else 
            return false;
    }
    int get_square_sum(int n)
    {
        int sum = 0;
        while(n != 0)
        {
            int single_num = n%10;
            sum += pow(single_num, 2);
            n /= 10;
        }
        return sum;
    }
    
};