難度: Easy
類型: Array,Matrix
CPP程式下載: 766.cpp
Topic:
Given an m x n matrix, return true if the matrix is Toeplitz. Otherwise, return false.
A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same elements.
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.
Constraints:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 200 <= matrix[i][j] <= 99
Follow up:
- What if the
matrixis 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
matrixis so large that you can only load up a partial row into the memory at once?
Consideration:
1. "Union of first column and first row elements".
2. Check each elements in the union from its left upper to right bottom diagonal direction. If each element's diagonal values are the same, then return true. If any different value is found, return false.
Code:
class Solution {
public:
bool isToeplitzMatrix(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
//cout << "m = " << m << ", n = "<< n << "\n";
//bool result = true;
int i,j,k,value;
for (k=0;k<m;k++)
{
j=0;
i=k;
value=matrix[i][j];
i++;
j++;
while (j<n && i<m)
{
//cout << "i = " << i << ", j = "<< j << "\n";
if (matrix[i][j]==value)
{
i++;
j++;
}
else
{
return false;
}
}
}
for (k=1;k<n;k++)
{
i=0;
j=k;
value=matrix[i][j];
i++;
j++;
while (j<n && i<m)
{
//cout << "i = " << i << ", j = "<< j << "\n";
if (matrix[i][j]==value)
{
i++;
j++;
}
else
{
return false;
}
}
}
return true;
}
};
Result:
Runtime
0msBeats100.00%
Memory
21.12MBBeats28.29%
