LeetCode--Happy Number

浏览数：109 / 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

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

本次题目主要是检查一个整数是否为Happy Number。一个好处是，输入的一定是正整数，无需考虑多种输入模式。主要思路：

1.首先要对输入的数分解，把它拆分成个、十、百、千位分开，便于下一步计算平方和；

2.结束条件：若该数值是Happy Number，则终止条件为某次循环中sum为1；若该数值不是Happy Number，则终止条件为一次循环，本次循环得到的sum是否在之前的循环中得到过，若得到过则该数一定不是Happy Number。

代码如下：

　　 public static ArrayList<Integer> l = new ArrayList<Integer>();
    public static void part(int n){
        if(n>=10)
            part(n/10);
        l.add(n%10);
        
    }
    
    
    public static boolean isHappy(int n) {
        ArrayList<Integer> h = new ArrayList<Integer>();
        h.add(n);
        int sum = 0;
        int m = n;
        while(sum!=1 && !h.contains(sum)){
            h.add(sum);
            sum = 0;
            l.clear();
            part(n);
            for(int i=0; i<l.size(); i++){
                sum = sum + l.get(i)*l.get(i);
            }
            n = sum;
        }
        if(sum == 1)
            return true;
        else
            return false;
    }