Leetcode解题: happy number (202)

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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

解法1：hashmap

这里需要判断重复的情况，想到用hashmap来存储已经计算过的值。对每一个未访问过的数值，计算位数的平方和。
C++

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class Solution {
public:
    bool isHappy(int n) {
        if (n <= 0) {
            return false;
        }
        if (n == 1) {
            return true;
        }
        unordered_map<int, bool> map;
        while (n != 1) {
            map[n] = true;
            int temp = digitSquareSum(n);
            if (map[temp]) {return false;}
            n = temp;
        }
        return true;
    }
    int digitSquareSum(int n) {
        if (n == 0 || n == 1) {
            return n;
        }
        int sum = 0;
        while (n != 0) {
            int digit = n % 10;
            sum += digit * digit;
            n /= 10;
        }
        return sum;
    }
};

Java

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解法2：观察规律

看到别人的解法里有这么一个巧妙的方法：可以试试几个会出现loop的数，最后都会出现4。结论是：所有最终会有4的都不是happy number，这样我们就可以把额外的空间要求去除了。
例子如下：
1^2 + 1^2 = 2
2^2 = 4
4^2 = 16
1^2 + 6^2 = 37
3^2 + 7^2 = 58
5^2 + 8^2 = 89
8^2 + 9^2 = 145
1^2 + 4^2 + 5^2 = 42
4^2 + 2^2 = 20
2^2 + 0^2 = 4

C++

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class Solution {
public:
    bool isHappy(int n) {
        while (n != 1 && n != 4) {
            int t = 0;
            while (n) {
                t += (n % 10) * (n % 10);
                n /= 10;
            }
            n = t;
        }
        return n == 1;
    }
};