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?
Time O(nm) space O(N+M)
public class Solution { public void setZeroes(int[][] matrix) { int col = matrix[0].length; int row = matrix.length; HashSet<Integer> row0 = new HashSet<>(); HashSet<Integer> col0 = new HashSet<>(); for(int i = 0 ; i < row ; i++){ for(int j = 0 ; j < col; j++){ if(matrix[i][j] == 0){ row0.add(i); col0.add(j); } } } for(int i : row0){ for(int j = 0 ; j < col; j++){ matrix[i][j] = 0; } } for(int i : col0){ for(int j = 0 ; j < row; j++){ matrix[j][i] = 0; } } } }
法二 : no space
用每行的第一个 和每列的第一来存0,注意清零时不能用第一列和第一行 否则会相互影响。
//No extra space 只要有0,用该row或者该col的第一个值存0 public void setZeroes(int[][] matrix) { int col = matrix[0].length; int row = matrix.length; int col0 = 1; for(int i = 0 ; i < row ; i++){ if (matrix[i][0] == 0) col0 = 0; for(int j = 1 ; j < col; j++){ if(matrix[i][j] == 0){ matrix[i][0] = 0; matrix[0][j] = 0; } } } for(int i = row -1 ; i >= 0 ; i--){ for(int j = col - 1 ; j >= 1; j--){ if(matrix[i][0] == 0 || matrix[0][j] == 0) matrix[i][j] = 0; } } if(col0 == 0){ for(int i = 0; i < row ; i++){ matrix[i][0] = 0; } } }