LeetCode 0202 Happy Number

LeetCode 0202 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.

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Example:

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

思路1

1. 获取每位数字，计算平方和
2. 使用哈希表记录，如不存在添加，若存在说明存在循环，退出，到达最后的结果1也退出

时间复杂度 $O(log N)$ 空间复杂度 $O(log N)$

代码1

class Solution {
public:
    bool isHappy(int n) {
        if(n < 1)  return false;
        unordered_set s;
        while(n != 1 && !s.count(n))
        {
            s.insert(n);
            int new_n = 0;
            while(n)
            {
                int t = n % 10;
                new_n += t * t;
                n /= 10;
            }
            n = new_n;
        }
        return n == 1;
    }
};

思路2

弗洛伊德循环查找算法: 隐式链表检测是否有环

1. 跟踪两个值，称为快跑者和慢跑者。在算法的每一步中，慢速在链表中前进 1 个节点，快跑者前进 2 个节点（对 getNext(n) 函数的嵌套调用）。
2. 如果 n 是一个快乐数，即没有循环，那么快跑者最终会比慢跑者先到达数字 1。
3. 如果 n 不是一个快乐的数字，那么最终快跑者和慢跑者将在同一个数字上相遇。

时间复杂度 $O(logN)$ 空间复杂度 $O(1)$

代码2

class Solution {
public:
    int next(int n)
    {
        int ans = 0;
        while(n)
        {
            int t = n%10;
            ans += t * t;
            n /= 10;
        }
        return ans;
    }
    
    bool isHappy(int n) {
        int slow = n;
        int fast = next(n);
        while(fast != 1 && slow != fast)
        {
            slow = next(slow);
            fast = next(next(fast));
        }
        return fast == 1;
    }
};