力扣每日一题2022-09-30中等题：零矩阵

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题目描述

零矩阵

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思路

模拟

记录矩阵中的0出现的横纵坐标，然后按横纵坐标修改矩阵中的值即可。

Python实现

class Solution:
    def setZeroes(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        m, n = len(matrix), len(matrix[0])
        row, col = [False] * m, [False] * n
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == 0:
                    row[i] = col[j] = True
        for i in range(m):
            for j in range(n):
                if row[i] or col[j]:
                    matrix[i][j] = 0

Java实现

class Solution {
    public void setZeroes(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        boolean[] row = new boolean[m];
        boolean[] col = new boolean[n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 0) {
                    row[i] = col[j] = true;
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (row[i] || col[j]) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
}

C++实现

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        vector<int> row(m), col(n);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (!matrix[i][j]) {
                    row[i] = col[j] = true;
                }
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (row[i] || col[j]) {
                    matrix[i][j] = 0;
                }
            }
        }
    }
};