LeetCode #905 SortArrayByParity. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Runtime: 32 ms , faster than 32.07% of C++ online submissions for Sort Array By Parity.
Memory Usage: 9.6 MB , less than 70.44% of C++ online submissions for Sort Array By Parity.
LeetCode #905
Q.
Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.
You may return any answer array that satisfies this condition.
주어진 양의 정수 배열 A 에서, 모든 짝수들이 배열 앞쪽으로 오고, 모든 홀수들이 뒤쪽으로 몰려있는 배열을 반환해라.
짝수와 홀수 위치만 맞으면 짝수 내, 홀수 내의 순서는 상관 없다.
Example 1:
Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Note:
1 <= A.length <= 5000
0 <= A[i] <= 5000
Process
// Process
//2. 시작부터 끝까지 반복한다.
// 2.1. 짝수인지 확인해서
// 2.1.1. 짝수면 - 짝수 마지막 index+1 번째 값과 교체한다.
//3. 배열 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
int idxLastEvenElem = 0;
int buff;
for (int i = 0; i < A.size(); ++i) {
if (isEven(A[i])) {
buff = A[idxLastEvenElem];
A[idxLastEvenElem] = A[i];
A[i] = buff;
++idxLastEvenElem;
}
}
return A;
}
private:
bool isEven(int elem) {
return elem % 2 == 0;
}
};
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