【Leetcode Algorithm】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.

Example: 19 is a happy number

• 1² + 9² = 82
• 8² + 2² = 68
• 6² + 8² = 100
• 1² + 0² + 0² = 1

第一次尝试代码（错误）：

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public class Solution {

    public boolean isHappy(int n) {

        int x = 0;

        //如果n为1,则直接返回true

        if(n==1){

            return true;

        }

        //计算各位数字的平方和

        while(n!=0){

            x += Math.pow((n%10),2);

            n = n/10;

        }

        //递归

        if(isHappy(x)){

            return true;

        }

        return false;

    }

}

出现无限循环，导致超时。（注意：Java求幂运算a^2，要用Math.pow(a,2)）

2——>4——>16——>37——>58——>89——>145——>42——>20——>4

所以，要避免循环，就要记录下计算出的每一个数，如果发现有重复的(用HashMap)，说明有循环。则判定为不是Happy Number

第二次尝试代码（正确）；

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public class Solution {

    public boolean isHappy(int n) {

        //HashMap用来判断是否有死循环

        HashMap hm = new HashMap();

        int x = n;

        while(true){

            if(x==1){

                return true;

            }

            //如果有死循环，则返回false

            if(hm.containsKey(x)){

                return false;

            }

            //将算出的新数放入HashMap

            hm.put(x,1);

            n = x;

            x = 0;

            //计算新数

            while(n!=0){

                x += Math.pow((n%10),2);

                n = n/10;

            }

             

        }

    }

}