给定一个 m x n
的矩阵,如果一个元素为 0 ,则将其所在行和列的所有元素都设为 0 。请使用 原地 算法。
进阶:
O(mn)
的额外空间,但这并不是一个好的解决方案。O(m + n)
的额外空间,但这仍然不是最好的解决方案。输入:matrix = [[1,1,1],[1,0,1],[1,1,1]]
输出:[[1,0,1],[0,0,0],[1,0,1]]
输入:matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]]
输出:[[0,0,0,0],[0,4,5,0],[0,3,1,0]]
提示:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-2^31 <= matrix[i][j] <= 2^31 - 1
0
public void setZeroes(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
boolean[] rows = new boolean[m];
boolean[] columns = new boolean[n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 0) {
rows[i] = columns[j] = true;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (rows[i] || columns[j]) {
matrix[i][j] = 0;
}
}
}
}
0
。public void setZeroes(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
boolean row0 = false, column0 = false;
for (int j = 0; j < n; j++) {
if (matrix[0][j] == 0) {
row0 = true;
break;
}
}
for (int i = 0; i < m; i++) {
if (matrix[i][0] == 0) {
column0 = true;
break;
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (matrix[i][j] == 0) {
matrix[i][0] = matrix[0][j] = 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 (row0) {
Arrays.fill(matrix[0], 0);
}
if (column0) {
for (int i = 0; i < m; i++) {
matrix[i][0] = 0;
}
}
}