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 };