Solution
unknown
javascript
4 years ago
778 B
5
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; } } };
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