108 lines
1.9 KiB
Markdown
108 lines
1.9 KiB
Markdown
# Leetcode Happy-Number
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#### 2022-07-26 09:12
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> ##### Algorithms:
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>
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> #algorithm #Floyd_s_cycle_finding_algorithm
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty_easy
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/happy-number/)
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---
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### Problem
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Write an algorithm to determine if a number `n` is happy.
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A **happy number** is a number defined by the following process:
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- Starting with any positive integer, replace the number by the sum of the squares of its digits.
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- Repeat the process until the number equals 1 (where it will stay), or it **loops endlessly in a cycle** which does not include 1.
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- Those numbers for which this process **ends in 1** are happy.
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Return `true` _if_ `n` _is a happy number, and_ `false` _if not_.
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#### Examples
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**Example 1:**
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**Input:** n = 19
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**Output:** true
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**Explanation:**
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12 + 92 = 82
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82 + 22 = 68
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62 + 82 = 100
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12 + 02 + 02 = 1
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**Example 2:**
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**Input:** n = 2
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**Output:** false
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#### Constraints
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- `1 <= n <= 231 - 1`
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### Thoughts
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> [!summary]
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> This is a #Floyd_s_cycle_finding_algorithm
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This works, because as the problem mentioned, this will result in a endless loop.
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So, by using fast ans slow, we can determine whether there is a loop.
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And, when fast hit 1, we know slow will eventually reach the
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answer, so we return early. (but in the cost of time of checking).
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### Solution
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```cpp
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class Solution {
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int getDigitSqrt(int i) {
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int sum = 0;
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while (i) {
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sum += (i % 10) * (i % 10);
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i /= 10;
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}
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return sum;
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}
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public:
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bool isHappy(int n) {
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// Floyd cycle finding algorighm.
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int slow, fast;
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slow = fast = n;
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do {
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slow = getDigitSqrt(slow);
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fast = getDigitSqrt(fast);
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fast = getDigitSqrt(fast);
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if (fast == 1) {
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return true;
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}
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} while (fast != slow);
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return false;
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}
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};
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```
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