Title Description

   Happy number Defined as ：

• For a positive integer , Replace the number with the sum of the squares of the numbers in each of its positions .
• Then repeat the process until the number becomes 1, It could be Infinite loop But it doesn't change 1.
• If this process The result is 1, So this number is the happy number .

If n yes Happy number Just go back to true ; No , Then return to false .

Example 1

 Input ：n = 19
 Output ：true

explain ：
1² + 9² = 82
8² + 2² = 68
6² + 8² = 100
1² + 0² + 0² = 1

Example 2

 Input ：n = 2
 Output ：false

I know you're too lazy to calculate and draw ,“ Uncle ” Spoil you . explain

Through the above figure and example 1 You may have a way of thinking , You can do it yourself first . Let's look at the following problem-solving ideas .

Their thinking

recursive
You can see that from the figure above , If it's not happy numbers , after n Time calculation , It is bound to enter a dead circle , That is, you will get the number repeatedly calculated before （ utilize Hashtable Solve this problem ）. The number of happiness is calculated a finite number of times 1. Use these two conditions as the conditions for the end of recursion , You can write recursion very quickly .

class Solution {
    
public:
	unordered_set<int> ans;// Use the hash table to determine whether the current value has been calculated 
	bool isHappy(int n) {
    
		int x=0;
		while(n){
    
			x+=pow(n%10,2);
			n=n/10;
		}
		if(x==1)// It's a happy number 
			return true;
		else{
    
			if(ans.count(x)==1)// This number has been calculated before , Not happy numbers 
				return false;
			else{
    
				ans.insert(x);
				return isHappy(x);
			}
		}
	}
};

Maybe some students will think of violence to solve , But we don't know what the maximum number of dead cycles will be , Unable to determine the number of cycles of the outer cycle . Personally, I don't recommend doing this , We need to know what a dead cycle stands for , Is to repeat the value calculated before . Use this as the recursive end condition , You can write recursion very quickly .
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