LeetCode 202. Happy Number

一、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

题意：给定一个数字，分别将数字拆为个位数，求其平方和，在拆分为个位数，再求其平方和，直到结果为1，那么n就是happy number

注意:结果为1返回true，那么什么时候返回false呢？就是当拆分个位数求平方和过程中出现无限循环？

        怎么样来判断进入死循环呢？就是每次将sum值载入到一个表中，当查询到当前sum值历史上已经出现过，直接返回false。

可以进行空间上的优化

class Solution {public:    bool isHappy(int n) {        //int count = 0;        map<int, int> mymap1;            while (true)    {                    int sum = 0;    while (n)         //将n拆分为个位数，同时求其平方和    {    //mymap.push_back(n % 10);    int temp = n%10;    sum += temp*temp;    n /= 10;    }    if (sum == 1)    {    //cout << "happy" << endl;    return true;    }    if(mymap1[sum])   //判断sum历史上出现过没    {        return false;    }    mymap1[sum]++;    //没出现过加入表中    n = sum;          //更新n进入下一轮循环       /* if (count++ > 10)    {    return false;    }*/    }    return false;    }};