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SetMatrixZeroes.java
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77 lines (75 loc) · 1.72 KB
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package com.yangchd.leetcode.medium;
import java.util.HashSet;
import java.util.Set;
/**
* @author yangchd 2018/10/31
*
* 73.Set Matrix Zeroes
* 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?
*
*/
public class SetMatrixZeroes {
/**
* 找出所有0所在的行和列，然后在赋值
*/
class Solution {
public void setZeroes(int[][] matrix) {
if (matrix.length == 0 || matrix[0].length == 0) {
return;
}
Set<Integer> x = new HashSet<Integer>();
Set<Integer> y = new HashSet<Integer>();
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == 0) {
x.add(i);
y.add(j);
}
}
}
for (int i : x) {
for (int m = 0; m < matrix[0].length; m++) {
matrix[i][m] = 0;
}
}
for (int j : y) {
for (int n = 0; n < matrix.length; n++) {
matrix[n][j] = 0;
}
}
}
}
}