Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
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?
解题思路:
This problem should be solved in place, i.e., no other array should be used. We can use the first column and the first row to track if a row/column should be set to 0.
Since we used the first row and first column to mark the zero row/column, the original values are changed.
Specifically, given, the following matrix
this problem can be solved by following 4 steps:
Step 1:
First row contains zero = true;
First column contains zero = false;
Step 2: use first row and column to mark zero row and column.
Step 3: set each elements by using marks in first row and column.
Step 4: Set first column and row by using marks in step 1.
Java code:
public class Solution { public void setZeroes(int[][] matrix) { if(matrix == null || matrix.length == 0 || matrix[0].length == 0) { return; } boolean firstrowzero = false; boolean firstcolzero = false; int m = matrix.length, n = matrix[0].length; //set first row and column zero or not for(int i = 0; i < m ; i++) { if(matrix[i][0] == 0) { firstcolzero = true; break; } } for(int j = 0; j < n ; j++) { if(matrix[0][j] == 0) { firstrowzero = true; break; } } // find zeroes and store the info in first row and column for(int i = 1; i< m ; i++) { for(int j = 1; j < n; j++) { if(matrix[i][j] == 0) { matrix[i][0] = 0; matrix[0][j] = 0; } } } //set zeroes except the first row and column 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; } } } // set zeroes for first row and column if needed if(firstrowzero){ for(int j = 0; j < n; j++) { matrix[0][j] = 0; } } if(firstcolzero){ for(int i = 0; i < m; i++) { matrix[i][0] = 0; } } } }
Reference:
1. http://www.programcreek.com/2012/12/leetcode-set-matrix-zeroes-java/