LeetCode #867 TransposeMatrix. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Runtime: 28 ms, faster than 58.04% of C++ online submissions for Transpose Matrix.
Memory Usage: 11.8 MB, less than 23.08% of C++ online submissions for Transpose Matrix.
LeetCode #867
Q.
Given a matrix A, return the transpose of A.
The transpose of a matrix is the matrix flipped over it's main diagonal, switching the row and column indices of the matrix.
매트릭스 A 가 주어지고, A 의 전치를 반환해라
행렬의 전치는, 행렬을 대각선으로 뒤집는 것인데, row 와 column 의 인덱스를 바꾼다.
Example 1:
Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: [[1,4,7],[2,5,8],[3,6,9]]
Example 2:
Input: [[1,2,3],[4,5,6]]
Output: [[1,4],[2,5],[3,6]]
Note:
- 1 <= A.length <= 1000
- 1 <= A[0].length <= 1000
Process
// Process
//1. Input 2d vector A
//2. Iterate first element size
// 2.1. Make vector
// 2.2. Iterate A size
// 2.2.1. Fill in vector
// 2.3. Add to resultVector
//3. Return resultVector
// 처리과정
//1. 2차원 배열 입력받는다.
//2. 2차원배열의 첫번째 값의 사이즈만큼 반복한다.
// 2.1. 공벡터 만들어둔다.
// 2.2. 2차원배열 사이즈만큼 반복한다.
// 2.2.1. 공벡터에 채워넣는다.
// 2.3. 결과벡터에 추가한다.
//3. 결과벡터 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
vector<vector<int>> transpose(vector<vector<int>>& A) {
vector<vector<int>> resultVector;
for (int i = 0; i < A[0].size(); ++i) {
vector<int> tempVec;
for (int j = 0; j < A.size(); ++j) {
tempVec.push_back(A[j][i]);
}
resultVector.push_back(tempVec);
}
return resultVector;
}
};
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