LeetCode(七)HashSet 202. Happy Number

202. Happy Number

题目要求如下

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

一开始没有理解题目的意思，原来是说如果一直没有1并且最终是无限循环下去。
解法如下

public class Solution {    public boolean isHappy(int n) {    Set<Integer> inLoop = new HashSet<Integer>();    int squareSum,remain;    while (inLoop.add(n)) {        squareSum = 0;        while (n > 0) {            remain = n%10;            squareSum += remain*remain;            n /= 10;        }        if (squareSum == 1)            return true;        else            n = squareSum;    }    return false;}}

使用HashSet的好处是可以检测是否已经开始无限循环了，只要有重复的数字就会跳出while循环。
使用HashMap可以参考LeetCode(三)