LeetCode 73. Set Matrix Zeroes

1. 题目描述

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

click to show follow up.

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?

2. 解题思路

我们这里使用两个set 分别记录 包含 0 的行 和 列的信息， 通过两次遍历实现题目要求

3. code

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        unordered_set<int> rows, cols;

        for (int i = 0; i != matrix.size(); i++){
            for (int j = 0; j != matrix[0].size(); j++){
                if (matrix[i][j] == 0){
                    rows.insert(i);
                    cols.insert(j);
                }
            }
        }

        for (auto iter = rows.begin(); iter != rows.end(); iter++){
            for (int i = 0; i != matrix[0].size(); i++){
                matrix[*iter][i] = 0;
            }
        }

        for (auto iter = cols.begin(); iter != cols.end(); iter++){
            for (int i = 0; i != matrix.size(); i++){
                matrix[i][*iter] = 0;
            }
        }
    }
};

4. 大神解法

这个解法， 实际上是将 数据覆盖到他的第 0 行 和 第 0 列进行相应处理。

/* My idea is simple: store states of each row in the first of that row, and store states of each column in the first of that column. Because the state of row0 and the state of column0 would occupy the same cell, I let it be the state of row0, and use another variable "col0" for column0. In the first phase, use matrix elements to set states in a top-down way. In the second phase, use states to set matrix elements in a bottom-up way. */

void setZeroes(vector<vector<int> > &matrix) {
    int col0 = 1, rows = matrix.size(), cols = matrix[0].size();

    for (int i = 0; i < rows; i++) {
        if (matrix[i][0] == 0) col0 = 0;
        for (int j = 1; j < cols; j++)
            if (matrix[i][j] == 0)
                matrix[i][0] = matrix[0][j] = 0;
    }

    for (int i = rows - 1; i >= 0; i--) {
        for (int j = cols - 1; j >= 1; j--)
            if (matrix[i][0] == 0 || matrix[0][j] == 0)
                matrix[i][j] = 0;
        if (col0 == 0) matrix[i][0] = 0;
    }
}