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: 1234 5123 9512 In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and 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]
.
解题思路:
乍一想可以从底部开始搜索,分两端搜索,一旦一斜行不相等即可返回false,但是这种方法代价太大。试一下:
class Solution {
public:
bool isToeplitzMatrix(vector<vector<int>>& matrix) {
if(matrix.size() == 1) return true;
int temp;
int temp_i,temp_j;
for(int i=matrix[0].size()-2;i>=0;i--){
temp = matrix[0][i];
temp_i = 0;
temp_j = i;
while(temp_j<matrix[0].size()-1&&temp_i<matrix.size()-1){
temp_i++;
temp_j++;
if(matrix[temp_i][temp_j] != temp) return false;
}
}
for(int i=matrix.size()-1;i>0;i--){
temp = matrix[i][0];
temp_i = i;
temp_j = 0;
while(temp_j<matrix[0].size()-1&&temp_i<matrix.size()-1){
temp_i++;
temp_j++;
if(matrix[temp_i][temp_j] != temp) return false;
}
}
return true;
}
};
可以跑了= =,不管了先。