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?
思路:主要考察的是空间复杂度,
方法一:用rowHasZero记录这一行是否有0 存在,用colHaszero记录这一列是否有0存在。后面处理rowHasZero和colHaszero。空间复杂度O(m+n)
class Solution { public: void setZeroes(vector<vector<int> > &matrix) { size_t row = matrix.size(); if(row == 0) return; size_t col = matrix[0].size(); vector<bool> rowHasZero(row, false); vector<bool> colHasZero(col, false); for(int i = 0; i < row; i++) { for(int j = 0; j < col; j++) { if(matrix[i][j] == 0) { rowHasZero[i] = true; colHasZero[j] = true; } } } for(int i = 0; i < row; i++) { if(rowHasZero[i]) { for(int j = 0; j < col; j++) { matrix[i][j] = 0; } } } for(int i = 0; i < col; i++) { if(colHasZero[i]) { for(int j = 0; j < row; j++) { matrix[j][i] = 0; } } } } };
方法二:空间复杂度O(1),用第0行记录列是否有0,用第0列记录行是否有0,先判断行0和列0是否有0,然后处理矩阵,若matrix[i][j]为0,设置matrix[i][0] = 0; matrix[0][j] = 0;,然后根据matrix[i][0]和matrix[0][j]来设置设置相应的行、列为0,最后处理行0,列0.
class Solution { public: void setZeroes(vector<vector<int> > &matrix) { size_t row = matrix.size(); if(row == 0) return; size_t col = matrix[0].size(); bool rowZeroHasZero = false; bool colZeroHasZero = false; for(int i = 0; i < col; i++) { if(matrix[0][i] == 0) { rowZeroHasZero = true; break; } } for(int i = 0; i < row; i++) { if(matrix[i][0] == 0) { colZeroHasZero = true; break; } } //cout << rowZeroHasZero << endl; //cout << colZeroHasZero << endl; for(int i = 1; i < row; i++) { for(int j = 1; j < col; j++) { if(matrix[i][j] == 0) { matrix[i][0] = 0; matrix[0][j] = 0; } } } for(int i = 1; i < row; i++) { if(matrix[i][0] == 0) { for(int j = 0; j < col; j++) { matrix[i][j] = 0; } } } for(int i = 1; i < col; i++) { if(matrix[0][i] == 0) { for(int j = 0; j < row; j++) { matrix[j][i] = 0; } } } //handle the first row & first col if(rowZeroHasZero) { for(int j = 0; j < col; j++) { matrix[0][j] = 0; } } if(colZeroHasZero) { for(int i = 0; i < row; i++) { matrix[i][0] = 0; } } } };