Solution
unknown
javascript
4 years ago
778 B
10
Indexable
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var setZeroes = function(matrix) {
let rows = [];
let columns = [];
for(let row = 0; row < matrix.length; row++) {
for(let column = 0; column < matrix[0].length; column++) {
if (matrix[row][column] === 0) {
rows.push(row);
columns.push(column);
}
}
}
for(let column of columns) {
for(let row = 0; row < matrix.length; row++) {
matrix[row][column] = 0;
}
}
for(let row of rows) {
for(let column = 0; column < matrix[0].length; column++) {
matrix[row][column] = 0;
}
}
};Editor is loading...