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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?
1 class Solution { 2 public: 3 void setZeroes(vector<vector<int>>& matrix) { 4 if (matrix.empty() || matrix[0].empty()) return; 5 // bool: rows[] stores the rows needed to be replaced by 0 6 // bool: cols[] stores the cols needed to be replaced by 0 7 int m = matrix.size(), n = matrix[0].size(); 8 vector<int> rows(m, false); 9 vector<int> cols(n, false); 10 for (int i = 0; i < m; i++) { 11 for (int j = 0; j < n; j++) { 12 if (!matrix[i][j]) { 13 rows[i] = true; 14 cols[j] = true; 15 } 16 } 17 } 18 19 // replace the matrix accordingly 20 for (int i = 0; i < m; i++) { 21 if (rows[i]) { 22 for (int j = 0; j < n; j++) 23 matrix[i][j] = 0; 24 } 25 } 26 for (int j = 0; j < n; j++) { 27 if (cols[j]) { 28 for (int i = 0; i < m; i++) 29 matrix[i][j] = 0; 30 } 31 } 32 } 33 };
Analyse: O(1) space.
-- Scan the matrix, when it comes to 0, set all 1 in its row and column as 2. Scan the matrix again, replace all 2 with 0. However, we may encounter the situation where 2 has already been used in the matrix...
-- Use the first row and first column to store the information. Solution should be as the same. A very smart solution can be viewed at:
1 class Solution { 2 public: 3 void setZeroes(vector<vector<int> > &matrix) { 4 int col0 = 1, rows = matrix.size(), cols = matrix[0].size(); 5 6 for (int i = 0; i < rows; i++) { 7 if (matrix[i][0] == 0) col0 = 0; 8 for (int j = 1; j < cols; j++) 9 if (matrix[i][j] == 0) 10 matrix[i][0] = matrix[0][j] = 0; 11 } 12 13 for (int i = rows - 1; i >= 0; i--) { 14 for (int j = cols - 1; j >= 1; j--) 15 if (matrix[i][0] == 0 || matrix[0][j] == 0) 16 matrix[i][j] = 0; 17 if (col0 == 0) matrix[i][0] = 0; 18 } 19 } 20 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/5883746.html
