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+# Leetcode Happy-Number
+
+#### 2022-07-26 09:12
+
+> ##### Algorithms:
+> #algorithm #Floyd_s_cycle_finding_algorithm 
+> ##### Difficulty:
+> #coding_problem #difficulty-easy
+> ##### Additional tags:
+> #leetcode 
+> ##### Revisions:
+> N/A
+
+##### Related topics:
+```expander
+tag:#Floyd_s_cycle_finding_algorithm 
+```
+ 
+ 
+
+##### Links:
+- [Link to problem](https://leetcode.com/problems/happy-number/)
+___
+### Problem
+
+Write an algorithm to determine if a number `n` 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.
+-   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.
+
+Return `true` _if_ `n` _is a happy number, and_ `false` _if not_.
+
+#### Examples
+
+**Example 1:**
+
+**Input:** n = 19
+**Output:** true
+**Explanation:**
+12 + 92 = 82
+82 + 22 = 68
+62 + 82 = 100
+12 + 02 + 02 = 1
+
+**Example 2:**
+
+**Input:** n = 2
+**Output:** false
+
+#### Constraints
+
+-   `1 <= n <= 231 - 1`
+
+### Thoughts
+
+> [!summary]
+> This is a #Floyd_s_cycle_finding_algorithm 
+
+This works, because as the problem mentioned, this will result in a endless loop.
+
+So, by using fast ans slow, we can determine whether there is a loop.
+
+And, when fast hit 1, we know slow will eventually reach the
+answer, so we return early. (but in the cost of time of checking).
+
+### Solution
+
+```cpp
+class Solution {
+  int getDigitSqrt(int i) {
+    int sum = 0;
+
+    while (i) {
+      sum += (i % 10) * (i % 10);
+      i /= 10;
+    }
+
+    return sum;
+  }
+
+public:
+  bool isHappy(int n) {
+    // Floyd cycle finding algorighm.
+    int slow, fast;
+    slow = fast = n;
+
+    do {
+      slow = getDigitSqrt(slow);
+      fast = getDigitSqrt(fast);
+      fast = getDigitSqrt(fast);
+      if (fast == 1) {
+        return true;
+      }
+    } while (fast != slow);
+
+    return false;
+  }
+};
+```
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