【leetcode100】矩阵置零

1、题目描述

给定一个 m x n 的矩阵，如果一个元素为 0 ，则将其所在行和列的所有元素都设为 0 。请使用原地算法。

示例 1：

输入：matrix = [[1,1,1],[1,0,1],[1,1,1]]
输出：[[1,0,1],[0,0,0],[1,0,1]]

2、初始思路

2.1 思路1

先找出所有0的横纵坐标，然后遍历置零。

class Solution(object):
    def setZeroes(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: None Do not return anything, modify matrix in-place instead.
        """
        all_i = []
        all_j = []
        m, n = len(matrix), len(matrix[0])
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == 0:
                    if i not in all_i:
                        all_i.append(i)
                    if j not in all_j:
                        all_j.append(j)
        #print(all_i)
        for i in all_i:
            for j in range(n):
                matrix[i][j] = 0
        for j in all_j:
            for i in range(m):
                matrix[i][j] = 0
        return matrix

 2.2 思路2

通过设置false来判断0的存在

class Solution(object):
    def setZeroes(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: None Do not return anything, modify matrix in-place instead.
        """
        m, n = len(matrix), len(matrix[0])
        m_0 = m * [False]
        n_0 = n * [False]
        for i in range(m):
            for j in range(n):
                if matrix[i][j] == 0:
                    m_0[i] = True
                    n_0[j] = True
        for i in range(m):
            for j in range(n):
                if m_0[i] or n_0[j]:
                    matrix[i][j] = 0

3、总结

1、矩阵的行列计算为

行
m = len(matrix)
列
n = len(matrix[0])

2、python中False和True首字母要大写