LeetCode之“数学”：Happy Number

　　题目要求：

　　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

　　Credits:
　　Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.

　　这道题比较需要注意的一点是要在合适的时候跳出死循环。程序如下：

 1 class Solution {
 2 public:
 3     bool isHappy(int n) {
 4         vector<int> vec;
 5         vec.push_back(n);
 6         while(n!= 1)
 7         {
 8             int sum = 0;
 9             int tmpN = n;
10             while(tmpN != 0)
11             {
12                 sum += (tmpN % 10) * (tmpN % 10);
13                 tmpN /= 10;
14             }
15             n = sum;
16             
17             if(find(vec.begin(), vec.end(), n) != vec.end())
18                 break;
19             else
20                 vec.push_back(n);
21         }
22         
23         return n == 1;
24     }
25 };

　　改为用哈希表来实现的程序如下：

 1 class Solution {
 2 public:
 3     bool isHappy(int n) {
 4         unordered_map<int, int> hashMap;
 5         hashMap[n] = n;
 6         while(n != 1)
 7         {
 8             int sum = 0;
 9             int tmpN = n;
10             while(tmpN != 0)
11             {
12                 sum += (tmpN % 10) * (tmpN % 10);
13                 tmpN /= 10;
14             }
15             n = sum;
16             
17             if(hashMap.find(n) != hashMap.end())
18                 break;
19             else
20                 hashMap[n] = n;
21         }
22         
23         return n == 1;
24     }
25 };