LeetCode #766 ToeplitzMatrix. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Runtime: 12 ms, faster than 88.34% of C++ online submissions for Toeplitz Matrix.
Memory Usage: 9.7 MB, less than 56.29% of C++ online submissions for Toeplitz Matrix.
LeetCode #766
Q.
A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same element.
Now given an M x N matrix, return True if and only if the matrix is Toeplitz.
Example 1:
Input: matrix = [ [1,2,3,4], [5,1,2,3], [9,5,1,2] ]
Output: True
Explanation: In the above grid, the diagonals are: "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]". In each diagonal all elements are the same, so the answer is True.
Example 2:
Input: matrix = [ [1,2], [2,2] ]
Output: False
Explanation: The diagonal "[1, 2]" has different elements.
Note:
- matrix will be a 2D array of integers.
- matrix will have a number of rows and columns in range [1, 20].
- matrix[i][j] will be integers in range [0, 99].
Follow up:
- What if the matrix is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once?
- What if the matrix is so large that you can only load up a partial row into the memory at once?
Process
// Process
//1. Input 2d matrix
//2. Iterate from begin to the width (i) (half right top on the matrix)
// 2.1. Iterate from begin to the height (j)
// 2.1.1. Check if [0][i] == [j][i+j]
// 2.1.1.1. If not -> false
//3. Iterate from 1 to the height (i) (half left bottom on the matrix)
// 3.1. Iterate from 1 to the height (j)
// 3.1.1. Check if [i][0] == [i+j][j]
// 3.1.1.1. If not -> false
//4. Return answer
// 처리과정
//1. 매트릭스 입력받는다.
//2. 우상단 자른 부분 확인한다.
//3. 좌하단 나머지 부분 확인한다.
//4. 결과 출력한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
bool isToeplitzMatrix(vector<vector<int>>& matrix) {
bool answer = true;
if (matrix[0].size() > 0) {
for (int i = 0; i < matrix[0].size(); ++i) {
for (int j = 0; j < matrix.size(); ++j) {
if (j+i < matrix[0].size())
if (matrix[0][i] != matrix[j][j+i]) {
answer = false;
}
}
}
if (answer) {
for (int i = 1; i < matrix.size() - 1; ++i) {
for (int j = 0; j < matrix[i].size(); ++j) {
if (j+i < matrix.size())
if (matrix[i][0] != matrix[i+j][j]) {
answer = false;
}
}
}
}
}
return answer;
}
};
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