leetcode笔记：Happy Number

一. 题目描述

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 （各个位的平方和）
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

二. 题目分析

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，若经过若干次运算后结果收敛到1，则该数字为Happy Number，不是Happy Number的数在经过多次运算后会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，并使用一个map存储已经出现过的数字，这样每轮计算完成后查找map，若发现值已存在，证明已陷入循环，可跳出循环并判定该整数不是Happy Number。

更多关于Happy Number的知识可参考：http://baike.baidu.com/link?url=slZGeshN-Igmd4geJ7PZ9hPwk_PZGZK6QttvzH5TpUiQIPy8qZAmG6o9-5-x-Eu2WGoQuhnASB7alb2ecEatpVj0C9-3DaQPjy0Cpvmvp7AB3IiKy6vu1Qb8FhCbj_Eg1Nt4ba9FDzZT1BzbQcsL5a

三. 示例代码

#include <iostream>
#include <map>

using namespace std;

class Solution
{
public:
    bool isHappy(int n)
    {
        if (n < 0) return false;
        if (n == 1) return true;

        map<int, bool> showNum;

        while (true)
        {
            int temp = 0;
            while (n)
            {
                temp += (n % 10) * (n % 10);
                n /= 10;
            }
            if (temp == 1)
                return true;
            else if (showNum[temp] == true) // 陷入循环，判为非快乐数
                return false;
            else
            {
                showNum[temp] = true;
                n = temp;
            }
        }
    }
};

一些测试结果：

四. 小结

无