Leetcode NO.202 Happy Number

本题题目要求如下：

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

• 12 + 92 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 = 1

其实这道题的思路并不难，只不过我久疏战阵，实在是做不出来了。。。。

这道题跟我原来做过的cycle-linkedlist有点像。。仔细分析题目的话，判断条件很简单，就是判断是否出线非1的无限循环。。。而只要是无限循环，就可以用我们之前用过的two pointers的方法解决：1个pointer慢（每次走一步），一个pointer快（每次走两步），如果是循环的话，这两个pointer就会再次相遇。。

代码如下：

class Solution {public:    bool isHappy(int n) {    int fast = cal_n(cal_n(n));    int slow = cal_n(n);        while (true) {        if (slow == 1)        return true;        else if (slow == fast)        return false;                slow = cal_n(slow);        fast = cal_n(cal_n(fast));        }    }private:int cal_n(int n) {int ret = 0;string s = to_string(n);for (int i = 0; i < s.size(); ++i) {ret += pow((s[i] - '0'), 2);}return ret;}};

再上传一个hashmap版本的，跟two pointer 基本思路一致。

class Solution {public:    unordered_set<int> hashset;    bool isHappy(int n) {        hashset.insert(n);        while (n != 1) {            n = cal(n);            if (hashset.find(n) != hashset.end() and n != 1) {                return false;            }            hashset.insert(n);        }        return true;    }private:    int cal(int n) {        int cnt = 0;        while (n > 0) {            cnt += pow((n % 10), 2);            n /= 10;        }        return cnt;    }};