73. Set Matrix Zeroes
Tags: ‘Array’
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?
Solution
方法1 Use extra two arrays to store where a column/ row has 0.
O(m * n) time O(m + n) space
方法2
Use row 0 and column 0 to store if current row/column has 0.
Finally iterate through matrix[i][j], if matrix[i][0] == 0 matrix[0][j] == 0, set matrix[i][j] 0.
O(m*n) time linear O(1) space
public void setZeroes(int[][] matrix) {
final int m = matrix.length;
final int n = matrix[0].length;
boolean row_has_zero = false; // 第一行是否存在 0
boolean col_has_zero = false; // 第一列是否存在 0
for (int i = 0; i < n; i++)
if (matrix[0][i] == 0) {
row_has_zero = true;
break;
}
for (int i = 0; i < m; i++)
if (matrix[i][0] == 0) {
col_has_zero = true;
break;
}
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
if (matrix[i][j] == 0) {
matrix[0][j] = 0;
matrix[i][0] = 0;
}
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
if (row_has_zero)
for (int i = 0; i < n; i++)
matrix[0][i] = 0;
if (col_has_zero)
for (int i = 0; i < m; i++)
matrix[i][0] = 0;
}