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SetMatrixZeroes.java
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117 lines (106 loc) · 2.59 KB
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package leetcode;
/**
* @author eko
* @date 2018/10/23 7:49 PM
*
* 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.
*/
public class SetMatrixZeroes {
public static void main(String[] args) {
int[][] matrix =
{
{1, 1, 1},
{1, 0, 1},
{1, 1, 1}
};
new SetMatrixZeroes().setZeroes(matrix);
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
System.out.printf(matrix[i][j] + "");
}
System.out.println();
}
}
public void setZeroes(int[][] matrix) {
if (matrix.length == 0) return;
boolean firstRow = false;
boolean firstCol = false;
for (int i = 0; i < matrix[0].length; i++) {
if (matrix[0][i] == 0) {
firstRow = true;
break;
}
}
for (int i = 0; i < matrix.length; i++) {
if (matrix[i][0] == 0) {
firstCol = true;
break;
}
}
for (int i = 1; i < matrix.length; i++) {
for (int j = 1; j < matrix[i].length; j++) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for (int i = 1; i < matrix.length; i++) {
if (matrix[i][0] == 0) {
for (int j = 1; j < matrix[i].length; j++) {
matrix[i][j] = 0;
}
}
}
for (int i = 1; i < matrix[0].length; i++) {
if (matrix[0][i] == 0) {
for (int j = 1; j < matrix.length; j++) {
matrix[j][i] = 0;
}
}
}
if (firstRow) {
for (int i = 0; i < matrix[0].length; i++) {
matrix[0][i] = 0;
}
}
if (firstCol) {
for (int i = 0; i < matrix.length; i++) {
matrix[i][0] = 0;
}
}
}
}