【leetcode】Happy Number（easy）

浏览数：29 / 时间：2015年06月11日

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

思路：

这题目挺有意思的，肯定不能真的通过无限循环来判断。我观察了一下，估计不happy的数字，运算一圈后会出现前面已经出现过的数字，即在一组数字里绕圈圈。故记录一下出现过的数字，判断是否重复。

如果重复了就不happy，得到1了就happy。

bool isHappy(int n) {
        unordered_set<int> record;
        while(1)
        {
            int sum = 0;
            while(n > 0)
            {
                sum += (n % 10) * (n % 10);
                n /= 10;
            }
            if(sum == 1)
                return true;
            
            if(record.find(sum) == record.end()) //当前数字没有出现过
            {
                record.insert(sum);
                n = sum;
            }
            else
                return false;

        }
    }

还有用O(1)空间的，就像链表找有没有圈一样，用快慢指针的思路。

public class Solution {
    public boolean isHappy(int n) {
        int x = n;
        int y = n;
        while(x>1){
            x = cal(x) ;
            if(x==1) return true ;
            y = cal(cal(y));
            if(y==1) return true ;

            if(x==y) return false;
        }
        return true ;
    }
    public int cal(int n){
        int x = n;
        int s = 0;
        while(x>0){
            s = s+(x%10)*(x%10);
            x = x/10;
        }
        return s ;
    }
}

还有用数学的，所有不happy的数字，都会得到4.（？？why）

bool isHappy(int n) {
    if (n <= 0) return false;

    int magic = 4;
    while (1) {
        if (n == 1) return true;
        if (n == magic) return false;
        int t = 0;
        while (n) {
            t += (n % 10) * (n % 10);
            n /= 10;
        }
        n = t;
    }
}