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 = 828² + 2² = 686² + 8² = 1001² + 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; 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; } };