Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
Follow up:
Did you use extra space?
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?
1.O(m+n)space,两个数组m和n分别存储第i行和第j列有没有零，比较慢
2.constant space ,利用矩阵的第零行和第零列分别存储第i行和第j列有没有零，需要注意的是再用两个变量row和column来表示第零行和第零列是否有零存在，因为这是边界情况
public class S073 {
public void setZeroes(int[][] matrix) {
//O(m+n) space slow
/* int [] m = new int[matrix.length];
int [] n = new int[matrix[0].length];
for (int i = 0;i<matrix.length;i++) {
for (int j = 0;j<matrix[0].length;j++) {
if (matrix[i][j] == 0) {
m[i] = -1;
n[j] = -1;
}
}
}
for (int i = 0;i<matrix.length;i++) {
if (m[i] == -1) {
for (int j = 0;j<matrix[0].length;j++) {
matrix[i][j] = 0;
if (n[j] == -1) {
for (int k = 0;k<matrix.length;k++) {
matrix[k][j] = 0;
}
}
}
}
} */
//利用矩阵的第零行和第零列分别来存每一列和每一行是否有零存在
int row = -1;
int column = -1;
for (int i = 0;i<matrix.length;i++) {
for (int j = 0;j<matrix[0].length;j++) {
if (matrix[i][j] == 0) {
if (i == 0 ) {
row = 0;
}
if (j == 0) {
column = 0;
}
matrix[0][j] = 0; //第零行
matrix[i][0] = 0; //第零列
}
}
}
for (int i = 1;i<matrix.length;i++) {
if (matrix[i][0] == 0) {
for (int j = 1;j<matrix[0].length;j++) {
matrix[i][j] = 0;
}
}
}
for (int j = 1;j<matrix[0].length;j++) {
if (matrix[0][j] == 0) {
for (int i = 1;i<matrix.length;i++) {
matrix[i][j] = 0;
}
}
}
if (row == 0) {
for (int i = 0;i<matrix[0].length;i++) {
matrix[0][i] = 0;
}
}
if (column == 0) {
for (int i = 0;i<matrix.length;i++) {
matrix[i][0] = 0;
}
}
}
}