LeetCode Set Matrix Zeroes
class Solution { public: void setZeroes(vector<vector<int> > &matrix) { int rows = matrix.size(); int cols = matrix[0].size(); bool col_has_zero = false; bool row_has_zero = false; for (int i=0; i<cols; i++) { if (matrix[0][i] == 0) { row_has_zero = true; break; } } for (int i=0; i<rows; i++) { if (matrix[i][0] == 0) { col_has_zero = true; break; } } for (int i=1; i<rows; i++) { for (int j=1; j<cols; j++) { if (matrix[i][j] != 0) continue; matrix[i][0] = 0; matrix[0][j] = 0; } } for (int i=1; i<rows; i++) { for (int j=1; j<cols; j++) { if (!matrix[i][0] || !matrix[0][j]) { matrix[i][j] = 0; } } } for (int i=0; row_has_zero && i<cols; i++) matrix[0][i] = 0; for (int i=0; col_has_zero && i<rows; i++) matrix[i][0] = 0; } };
没意思,搞个几个bit的额外空间会死么
第二轮:
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?
class Solution { public: void setZeroes(vector<vector<int> > &matrix) { bool first_row_zero = false; bool first_col_zero = false; int rows = matrix.size(); if (rows < 1) { return; } int cols = matrix[0].size(); if (cols < 1) { return; } for (int i=0; i<cols; i++) { if (matrix[0][i] == 0) { first_row_zero = true; break; } } for (int i=0; i<rows; i++) { if (matrix[i][0] == 0) { first_col_zero = true; break; } } for (int i=1; i<rows; i++) { for (int j=1; j<cols; j++) { if (matrix[i][j] == 0) { matrix[0][j] = 0; matrix[i][0] = 0; } } } for (int i=1; i<rows; i++) { if (matrix[i][0] == 0) { for (int j=1; j<cols; j++) matrix[i][j] = 0; } } for (int i=1; i<cols; i++) { if (matrix[0][i] == 0) { for (int j=1; j<rows; j++) matrix[j][i] = 0; } } for (int i=0; i<cols && first_row_zero; i++) matrix[0][i] = 0; for (int i=0; i<rows && first_col_zero; i++) matrix[i][0] = 0; } };
没有一次AC,最后做填充的时候应该出去最左边的列和最上面的行