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.
click to show follow up.
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?
SOLUTION 1:
题目要求O(1)的空间消耗。
1. 我们可以使用第一行,第一列来作为Flag,记录某一行,某一列是否应该被设置为0.
2. 因为第一行,第一列共用左上角的flag,所以我们需要另外找2个flag来定义第一行,第一列本身是否应该设置为0.
row1Zero, col1Zero
3. 先扫描首行,首列把首行首列的flag算出。
4. 扫描其他的矩阵,将第一行每一列的flag算出。
5. 设置矩阵中除了首行首列的cells.
6. 设置首行,设置首列。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 public void setZeroes(int[][] matrix) { 3 if (matrix == null || matrix.length == 0 4 || matrix[0].length == 0) { 5 return; 6 } 7 8 boolean row1Zero = false; 9 boolean col1Zero = false; 10 11 int rows = matrix.length; 12 int cols = matrix[0].length; 13 14 // Determine if the first column should be Zero. 15 for (int i = 0; i < rows; i++) { 16 if (matrix[i][0] == 0) { 17 col1Zero = true; 18 break; 19 } 20 } 21 22 // Determine if the first row should be Zero. 23 for (int i = 0; i < cols; i++) { 24 if (matrix[0][i] == 0) { 25 row1Zero = true; 26 break; 27 } 28 } 29 30 // we use the first row and the first col as the flag to record the 31 // cells whether or not set to 0. 32 for (int i = 1; i < rows; i++) { 33 for (int j = 1; j < cols; j++) { 34 // 注意了,这个矩阵是0和非0,并不是0和1. 35 if (matrix[i][j] == 0) { 36 // set the flag in the first line and the first column 37 matrix[i][0] = 0; 38 matrix[0][j] = 0; 39 } 40 } 41 } 42 43 // set the inner cells. 44 // Be careful: i, j start from 1. 45 for (int i = 1; i < rows; i++) { 46 for (int j = 1; j < cols; j++) { 47 if (matrix[i][0] == 0 48 || matrix[0][j] == 0) { 49 matrix[i][j] = 0; 50 } 51 } 52 } 53 54 // set the first row. 55 if (row1Zero) { 56 for (int i = 0; i < cols; i++) { 57 matrix[0][i] = 0; 58 } 59 } 60 61 // set the first col. 62 if (col1Zero) { 63 for (int i = 0; i < rows; i++) { 64 matrix[i][0] = 0; 65 } 66 } 67 68 return; 69 } 70 }
2014.1230 redo:
把前面三个循环合并,会简单一点儿。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 public void setZeroes(int[][] matrix) { 3 if (matrix == null || matrix.length == 0 || matrix[0].length == 0) { 4 return; 5 } 6 7 boolean row1 = false; 8 boolean col1 = false; 9 10 int rows = matrix.length; 11 int cols = matrix[0].length; 12 13 // set flags. 14 for (int i = 0; i < rows; i++) { 15 for (int j = 0; j < cols; j++) { 16 if (matrix[i][j] != 0) { 17 continue; 18 } 19 20 // set the flag of a column and a row. 21 matrix[0][j] = 0; 22 matrix[i][0] = 0; 23 24 // get flag of first row. 25 if (i == 0) { 26 row1 = true; 27 } 28 29 // get flag of first column. 30 if (j == 0) { 31 col1 = true; 32 } 33 } 34 } 35 36 // set the matrix. 37 for (int i = 1; i < rows; i++) { 38 for (int j = 1; j < cols; j++) { 39 if (matrix[0][j] == 0 || matrix[i][0] == 0) { 40 matrix[i][j] = 0; 41 } 42 } 43 } 44 45 // set first column 46 if (col1) { 47 for (int i = 0; i < rows; i++) { 48 // bug 1: can't use matrix[i][j] 49 matrix[i][0] = 0; 50 } 51 } 52 53 // set first row. 54 if (row1) { 55 for (int i = 0; i < cols; i++) { 56 matrix[0][i] = 0; 57 } 58 } 59 } 60 }
GitHub Code: