LeetCode #202 HappyNumber. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
30분정도.. 3번 제출.. 퍼포먼스 100%..
30min.. submit 3 times.. 100% performance..
LeetCode #202
Q.
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.
숫자가 "Happy" 한지 확인하는 알고리즘을 코딩해봐라
happy 한 숫자는 다음과 같은 프로세스로 정의된다: 양의 정수로 시작하고, 숫자들을 각 자릿수 제곱들의 합으로 바꾸면서, 그 합이 1이 될 때까지 반복한다. 아니면 무한대로 1이 안나오고 반복하게 될 수도 있다. 1이 나와서 끝나는 숫자를 happy한 숫자라고 한다.
e.g.
Example:
Input: 19
Output: true
Explanation: 설명
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1
Process
// Process
//1. Input integer
//2. Prepare previousIntList
//3. Iterate till Integer is 1 or isDone
// 3.1. Get sum of all each power of digits
//4. Return result
// 처리과정
//1. 정수 입력 받는다.
//2. 이전에 나온 정수 vector 를 준비한다.
//3. 정수가 1 이거나 inDone 일 때까지 반복한다.
// 3.1. 각 자릿수의 제곱 합을 구한다.
// 3.2. 이전에 나온 수인지 체크한다.
//4. 결과 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool isHappy(int n) {
vector<int> previousInt;
bool isDone = false;
while (n != 1 && !isDone) {
sumOfAllPowerOfDigits(n);
if(std::find(previousInt.begin(), previousInt.end(), n) != previousInt.end()) {
isDone = true;
} else {
previousInt.push_back(n);
}
}
return n == 1;
}
private:
void sumOfAllPowerOfDigits(int& n) {
int sum = 0;
while (n != 0) {
sum += pow((n % 10), 2);
n /= 10;
}
n = sum;
}
};
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