LeetCode #258 AddDigits. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접
퍼포먼스는 100% 잘나왔고, 메모리는 안좋음
Performance is the best, memory is not good
LeetCode #171
Q.
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
양의정수 num 이 주어지고, 한자리수가 남을 때까지 반복적으로 각 자리수의 합을 구하라.
e.g.
Example:
Input: 38
Output: 2
Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.
Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
추가로 :
반복문이나 재귀 없이 시간복잡도 O(1) 에 해결 가능한가?
Process
// Process
//1. Input number
//2. Iterate while digitSum is not a single digit
// 2.1. Make digitSum
//3. Return digitSum
//makeDigitSum
//1. Input number
//2. Iterate while it's not a single digit
// 2.1. add remainder after divide by 10 to digitSum
//3. Return digitSum
// 처리과정
//1. 숫자를 입력받는다.
//2. digitSum 이 한자리수가 아닐 때까지 반복한다.
// 2.1. digitSum 을 만든다.
//3. digitSum 반환한다.
//makeDigitSum
//1. 숫자를 입력받는다.
//2. 한자리수가 아니면 반복한다.
// 2.1. 10으로 나눈 나머지를 digitSum에 더한다.
//3. digitSum 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
#include <iostream>
using namespace std;
class Solution {
public:
int addDigits(int num) {
while (num > 9) {
num = makeDigitSum(num);
}
return num;
}
private:
int makeDigitSum(int number) {
int digitSum = 0;
while (number > 9)
{
digitSum += (number % 10);
number /= 10;
}
digitSum += number;
return digitSum;
}
};
int main(int argc, char *argv[]) {
Solution sln;
cout << sln.addDigits(38) << endl;
}
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