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set-matrix-zeroes.js
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set-matrix-zeroes.js
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// 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?
//
//
/**
* @param {number[][]} matrix
* @return {void} Do not return anything, modify matrix in-place instead.
*/
var setZeroes = function(matrix) {
var row = false, col = false;
var m = matrix.length, n = matrix[0].length;
var i, j;
for(i = 0; i < m; ++i){
for(j = 0; j < n; ++j){
if(matrix[i][j] === 0) {
if(i === 0) {
row = true;
}
if(j === 0){
col = true;
}
matrix[0][j] = matrix[i][0] = 0;
}
}
}
for(i = 1; i < m; i++){
for(j = 1; j < n; j++){
if(matrix[i][0] === 0 || matrix[0][j] === 0) matrix[i][j] = 0;
}
}
if(col){
for(i = 0; i < m; i++) {
matrix[i][0] = 0;
}
}
if(row){
for(j = 0; j < n; j++) {
matrix[0][j] = 0;
}
}
};