202. Happy Numbe

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

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

这道题和 258.Add Dighits 的第一种解法很像，只不过上一道题的终止条件是小于10，而这道题的终止条件是无限循环（loops endlessly），也就是再循环过程中出现了以前出现的数字，所以这道题就简化成：判断是否出现过这个数字。

通常采用的方法是：

• 创建一个set或者hash_set函数，在这里我用的是CODEBLOCKS，如果用hash_set，会出现 error: 'hash_set' was not declared in this scope 的错误，VS2013上可以。

在这里我用的是set

#include <set>
#include <iostream>

using namespace std;

    bool isHappy(int n) {

        set<int> set;

        while(n!=1){
            int sum = 0;
            while(n>0){
                sum += (n % 10) * (n % 10);
                n = n / 10;
            }

            if(set.find(sum) != set.end()) //循环终止条件
            {
                cout<<2<<endl;
                return false;
            }
            else
                {
                    set.insert(sum);
                }

            n = sum;
        }
        cout<<1<<endl;
        return true;
    }

int main()
{
    int n;
    cin>>n;
    isHappy(n);

    return 0;
}

• 然后就是大神级别的方法了，在DISSCUS中看到如下方法，复杂度只有o(1)哦

int digitSquareSum(int n) {
    int sum = 0, tmp;
    while (n) {
        tmp = n % 10;
        sum += tmp * tmp;
        n /= 10;
    }
    return sum;
}

bool isHappy(int n) {
    int slow, fast;
    slow = fast = n;
    do {
        slow = digitSquareSum(slow);
        fast = digitSquareSum(fast);
        fast = digitSquareSum(fast);
    } while(slow != fast);
    if (slow == 1) return 1;
    else return 0;
}