202. Happy Number*

https://leetcode.com/problems/happy-number/

题目描述

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: 
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

C++ 实现 1

基本逻辑很简单, 主要需要考虑无限循环的问题, 比如: 89:

8^2 + 9^2 = 145
1^2 + 4^2 + 5^2 = 42
4^2 + 2^2 = 20
2^2 + 0^2 = 4
4^2       = 16
1^2 + 6^2 = 37
3^2 + 7^2 = 58
5^2 + 8^2 = 89

也就是说, 执行基本逻辑的情况下, 最后的结果又等于了初始数据 (89), 这样造成了无限循环. 因此, 需要使用哈希表记录下每个访问过的数据.

class Solution {
private:
    unordered_set<int> record;
public:
    bool isHappy(int n) {
        int sum = 0;
        while (true) {
            if (record.count(n)) return false;
            record.insert(n);
            while (n) {
                sum += (n % 10) * (n % 10);
                n /= 10;
            }
            if (sum == 1) break;
            n = sum;
            sum = 0;
        }
        return true;
    }
};

C++ 实现 2

使用快慢指针, 参考: My solution in C( O(1) space and no magic math property involved )

class Solution {
private:
    int digitSquareSum(int n) {
        int sum = 0, tmp;
        while (n) {
            tmp = n % 10;
            sum += tmp * tmp;
            n /= 10;
        }
        return sum;
    }
public:
    bool isHappy(int n) {
        int slow, fast;
        slow = fast = n;
        do {
            slow = digitSquareSum(slow);
            fast = digitSquareSum(fast);
            fast = digitSquareSum(fast);
        } while(slow != fast);
        return slow == 1;
    }
};