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?
Analysis:
O(m+n) space is easy. The hard problem is O(1) space solution.
O(m+n) space solution:
1 public class Solution { 2 public void setZeroes(int[][] matrix) { 3 int xLen = matrix.length; 4 if (xLen==0) return; 5 int yLen = matrix[0].length; 6 if (yLen==0) return; 7 8 boolean[] rowZero = new boolean[xLen]; 9 boolean[] colZero = new boolean[yLen]; 10 Arrays.fill(rowZero,false); 11 Arrays.fill(colZero,false); 12 13 for (int i=0;i<xLen;i++) 14 for (int j=0;j<yLen;j++) 15 if (matrix[i][j]==0){ 16 rowZero[i] = true; 17 colZero[j] = true; 18 } 19 20 for (int i=0;i<xLen;i++) 21 for (int j=0;j<yLen;j++) 22 if (rowZero[i] || colZero[j]) 23 matrix[i][j] = 0; 24 return; 25 } 26 }
O(1) space solution:
We can use the first row and first col to store the information whether the row and the col should be turn to zero. If m[i][j]==0, then we set m[0][j]=m[i][0]=0. After scan all matrix, we then turn to zeroes according the information.
At last, if the first row and first col have zero at the begnning, we further turn them into zeroes.
NOTE: This very smart solution is not figured out by mysefl, so READ IT AGAIN.
1 public class Solution { 2 public void setZeroes(int[][] matrix) { 3 int xLen = matrix.length; 4 if (xLen==0) return; 5 int yLen = matrix[0].length; 6 if (yLen==0) return; 7 8 boolean firstRowZero = false, firstColZero = false; 9 for (int i=0;i<xLen;i++) 10 if (matrix[i][0]==0) { 11 firstColZero = true; 12 break; 13 } 14 15 for (int i=0;i<yLen;i++) 16 if (matrix[0][i]==0) { 17 firstRowZero = true; 18 break; 19 } 20 21 for (int i=1;i<xLen;i++) 22 for (int j=1;j<yLen;j++) 23 if (matrix[i][j]==0){ 24 matrix[i][0] = 0; 25 matrix[0][j] = 0; 26 } 27 28 for (int i=1;i<xLen;i++) 29 for (int j=1;j<yLen;j++) 30 if (matrix[i][0]==0 || matrix[0][j]==0) 31 matrix[i][j] = 0; 32 if (firstRowZero) 33 for (int i=0;i<yLen;i++) matrix[0][i]=0; 34 if (firstColZero) 35 for (int i=0;i<xLen;i++) matrix[i][0]=0; 36 37 38 return; 39 } 40 }
原文：http://www.cnblogs.com/lishiblog/p/4102674.html