leetcode-【簡單題】Happy Number

標籤：

題目：

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

• 1^2 + 9^2 = 82
• 8^2 + 2^2 = 68
• 6^2 + 8^2 = 100
• 1^2 + 0^2 + 0^2 = 1

連結：

https://leetcode.com/problems/happy-number/

答案：

暴力解決，直接給每個數的每位計算平方和就行，然後把中間的數都記錄下來，如果下次出現這個數，就是loop了。

代碼：

 1 #include<vector> 2  3 using std::vector; 4  5 class Solution { 6 private: 7     vector<int> med; 8      9     bool inMed(int n)10     {11         vector<int>::iterator beg;12         for(beg = med.begin(); beg != med.end(); ++ beg)13         {14             if(*beg == n)15             {16                 return true;17             }18         }19         20         return false;21     }22     23     int sum(int n)24     {25         int s = 0;26         while(n != 0)27         {28             int modn = n % 10;29             s += modn * modn;30             n = n /10;31         }32         33         return s;34     }35     36 public:37     bool isHappy(int n) 38     {39         if(n == 1 || n == 19 || n == 13 || n == 10)40         {41             return true;42         }43         44         med.clear();45         med.push_back(n);46         47         int result = n;48         bool isLoop = false;49         while( result != 1)50         {51             result = sum(result);52             if(result != 1 && inMed(result))53             {54                 isLoop = true;55                 break;56             }57             58             med.push_back(result);59         }60         61         return !isLoop;62     }63 };

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代碼中的isHappy()方法的最開頭我是隨便加了幾個判斷，哈哈，題目中給出的快樂數應該會在背景測試資料中。

leetcode-【簡單題】Happy Number