[LeetCode]: 202: Happy Number

题目：

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

分析：

这道题的难点还是读题呀，又没有看懂题目的需求

题目要求对任意一个正整数，不断各个数位上数字的平方和，若最终收敛为1，则该数字为happy number，否则程序可能从某个数开始陷入循环。

这道题目我们只用根据规则进行计算，使用Hashset来存储结果，这样就可以判断结果是不是陷入循环了

 

代码：

    public boolean isHappy(int n) {  
        if(n<=0){
            return false;  
        } 
        long lnResult = n;  
        Set<Long> set = new HashSet<Long>();  
          
        while(lnResult<=Integer.MAX_VALUE) {  

            lnResult = digitSquare(lnResult);  
            if(lnResult == 1){
                return true;  
            } 
            
            if(set.contains(lnResult) ){
                return false;
            } else{
                set.add(lnResult);  
            }
        }  
        return false;  
    }  
      
    private long digitSquare(long lnInput) {  
        long Sum = 0;  
        while(lnInput!=0) {  
            Sum += Math.pow(lnInput%10, 2);  
            lnInput /= 10;  
        }  
        return Sum;  
    }