A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?
Solutions:
class Solution {
public void setZeroes(int[][] matrix) {
int row = matrix.length;
int col = matrix[0].length;
boolean fc = false, fr = false;
for (int i = 0; i < row; i++)
{
for (int j = 0; j < col; j++)
{
if (matrix[i][j] == 0)
{
if (i == 0) fr =true;
if (j == 0) fc = true;
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for (int i = 1; i < row; i++)
{
for (int j = 1; j < col; j++)
{
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
}
if(fr) {
for(int j = 0; j < matrix[0].length; j++) {
matrix[0][j] = 0;
}
}
if(fc) {
for(int i = 0; i < matrix.length; i++) {
matrix[i][0] = 0;
}
}
}
}
Points:
We use the corresponding element in the first column or row to mark which rows and columns we would like to set to 0 in the coming iterations. Problems will happen when the original 0 is in the first column or row, then we set a boolean to operate separately.
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
Example 1:
Input: [ [1,1,1], [1,0,1], [1,1,1] ] Output: [ [1,0,1], [0,0,0], [1,0,1] ] Example 2:
Input: [ [0,1,2,0], [3,4,5,2], [1,3,1,5] ] Output: [ [0,0,0,0], [0,4,5,0], [0,3,1,0] ] Follow up:
A straight forward solution using O(mn) space is probably a bad idea. A simple improvement uses O(m + n) space, but still not the best solution. Could you devise a constant space solution?
Solutions: class Solution { public void setZeroes(int[][] matrix) { int row = matrix.length; int col = matrix[0].length; boolean fc = false, fr = false; for (int i = 0; i < row; i++) { for (int j = 0; j < col; j++) { if (matrix[i][j] == 0) { if (i == 0) fr =true; if (j == 0) fc = true; matrix[i][0] = 0; matrix[0][j] = 0; } } } for (int i = 1; i < row; i++) { for (int j = 1; j < col; j++) { if (matrix[i][0] == 0 || matrix[0][j] == 0) matrix[i][j] = 0; } } if(fr) { for(int j = 0; j < matrix[0].length; j++) { matrix[0][j] = 0; } } if(fc) { for(int i = 0; i < matrix.length; i++) { matrix[i][0] = 0; } }
} }
Points: