LeetCode 73. Set Matrix Zeroes

Solution1：我的答案
比较笨，算法时间复杂度是O(mn)O(mn)O(mn)，占用额外空间是O(m+n)O(m+n)O(m+n)

class Solution {
public:void setZeroes(vector<vector<int>>& matrix) {if(matrix.size() * matrix[0].size() == 0)return;vector<int> row;    //某行有0则为truevector<int> col;    //某列有0则为truefor (int i = 0; i < matrix.size(); i++) { for (int j = 0; j < matrix[0].size(); j++) {if (matrix[i][j] == 0) {if(find(row.begin(), row.end(), i) == row.end())row.push_back(i);if(find(col.begin(), col.end(), j) == col.end())col.push_back(j);}}}for (int i = 0; i < row.size(); i++) for (int j = 0; j < matrix[i].size(); j++)matrix[row[i]][j] = 0;for (int j = 0; j < col.size(); j++) for (int i = 0; i < matrix.size(); i++)matrix[i][col[j]] = 0;return;}
};

Solution2：空间复杂度优化
利用第一行第一列做记录
比较笨，算法时间复杂度是O(mn)O(mn)O(mn)，占用额外空间是O(1)O(1)O(1)

class Solution {
public:void setZeroes(vector<vector<int>>& matrix) {int m = matrix.size(), n = matrix[0].size();bool row_zero = false, col_zero = false;for (int i = 0; i < n; i++) {if (matrix[0][i] == 0) {row_zero = true;break;}   }for (int i = 0; i < m; i++) {if (matrix[i][0] == 0) {col_zero = true;break;} }for (int i = 1; i < m; i++) { for (int j = 1; j < n; j++) {if (matrix[i][j] == 0) {matrix[i][0] = 0;matrix[0][j] = 0;}}}for (int i = 1; i < m; i++)for (int j = 1; j < n; j++)if(matrix[i][0] == 0 || matrix[0][j] == 0)matrix[i][j] = 0;if (row_zero) {for (int i = 0; i < n; i++)matrix[0][i] = 0;}if (col_zero) {for (int i = 0; i < m; i++)matrix[i][0] = 0;}return;}
};

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