Leetcode 73: set matrix zeros

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

Solution1:
O(n) 空间复杂度： 先扫描一遍matrix，记录存在0的行和列，最后再将这些行和列进行赋值0。
class Solution {
private:
    void setRow(vector<vector<int>>& matrix, int row) {
        for (int i = 0; i < matrix[0].size(); i++) {
            matrix[row][i] = 0;
        }
    }
    void setCol(vector<vector<int>>& matrix, int col){
        for (int i = 0; i < matrix.size();i++) {
            matrix[i][col] = 0;
        }
    }
public:
    void setZeroes(vector<vector<int>>& matrix) {
        //sanity check
        if (matrix.size() == 0 || matrix[0].size() == 0) return;
        set<int> col;
        set<int> row;
        for (int i = 0; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[0].size(); j++) {
                if (matrix[i][j] == 0) {
                    if (col.find(j) == col.end()) {
                        col.insert(j);
                    }
                    if (row.find(i) == row.end()) {
                        row.insert(i);
                    }
                }
            }
        }
        for (set<int> :: iterator it = col.begin(); it != col.end(); it++){
            setCol(matrix, *it);
        }
        for (set<int> :: iterator it = row.begin(); it != row.end(); it++) {
            setRow(matrix, *it);
        }
        return;
    }
};

Solution2:
对空间复杂度可以进一步优化: O(1) 利用当前matrix的第一行和第一列来记录置0的情况，如果扫描到一个0，就把当前的第一行上的那列置0，把当前的第一列的那行置0，最后再扫描一遍，将所有第一列和第一行的所有0所在列和行置0.对于第一列和第一行可以利用两个变量来看是否要全置0，这样做的空间复杂度为O（1）
class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        //sanity check
        if (matrix.size() == 0 && matrix[0].size() == 0) {
            return;
        }
        bool row = false;
        bool col = false;
        for (int i = 0; i < matrix.size(); i++) {
            if (matrix[i][0] == 0) {
                col = true;
                break;
            }
        }
        for (int i = 0; i < matrix[0].size(); i++) {
            if (matrix[0][i] == 0){
                row = true;
                break;
            }
        }
        for (int i = 1; i < matrix.size(); i++) {
            for (int j = 1; j < matrix[0].size(); j++) {
                if (matrix[i][j] == 0) {
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        for (int i = 1; i < matrix.size(); i++) {
            for (int j = 1; j < matrix[0].size(); j++) {
                if (matrix[0][j] == 0 || matrix[i][0] == 0) {
                    matrix[i][j] = 0;
                }
            }
        }
        if (col) {
            for (int i = 0; i < matrix.size(); i++){
                matrix[i][0] = 0;
            }
        }
        if (row) {
            for (int i = 0; i < matrix[0].size(); i++) {
                matrix[0][i] = 0;
            }
        }
        return;
    }
};