Python矩阵置零算法全解析

Python 矩阵置零问题解析

矩阵置零问题要求将矩阵中某个元素为零的行和列全部置零。这是一个经典的算法问题，常用于考察对矩阵操作的理解和实现能力。

问题描述

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。要求原地修改矩阵，即不使用额外的存储空间。

方法一：使用标记数组

通过两个标记数组分别记录需要置零的行和列。遍历矩阵，记录所有零元素的行和列。再次遍历矩阵，根据标记数组将对应行和列置零。

def setZeroes(matrix):
    m, n = len(matrix), len(matrix[0])
    rows, cols = [False] * m, [False] * n
    
    for i in range(m):
        for j in range(n):
            if matrix[i][j] == 0:
                rows[i] = True
                cols[j] = True
    
    for i in range(m):
        for j in range(n):
            if rows[i] or cols[j]:
                matrix[i][j] = 0

方法二：原地标记

利用矩阵的第一行和第一列作为标记空间。首先检查第一行和第一列是否需要置零。遍历矩阵其余部分，将零元素的位置记录在第一行和第一列。根据标记置零。

def setZeroes(matrix):
    m, n = len(matrix), len(matrix[0])
    first_row_zero = any(matrix[0][j] == 0 for j in range(n))
    first_col_zero = any(matrix[i][0] == 0 for i in range(m))
    
    for i in range(1, m):
        for j in range(1, n):
            if matrix[i][j] == 0:
                matrix[i][0] = 0
                matrix[0][j] = 0
    
    for i in range(1, m):
        for j in range(1, n):
            if matrix[i][0] == 0 or matrix[0][j] == 0:
                matrix[i][j] = 0
    
    if first_row_zero:
        for j in range(n):
            matrix[0][j] = 0
    if first_col_zero:
        for i in range(m):
            matrix[i][0] = 0

方法三：优化空间复杂度

进一步优化空间复杂度，仅使用一个变量标记第一列是否需要置零。其余步骤与方法二类似。

def setZeroes(matrix):
    m, n = len(matrix), len(matrix[0])
    first_col_zero = False
    
    for i in range(m):
        if matrix[i][0] == 0:
            first_col_zero = True
        for j in range(1, n):
            if matrix[i][j] == 0:
                matrix[i][0] = 0
                matrix[0][j] = 0
    
    for i in range(1, m):
        for j in range(1, n):
            if matrix[i][0] == 0 or matrix[0][j] == 0:
                matrix[i][j] = 0
    
    if matrix[0][0] == 0:
        for j in range(n):
            matrix[0][j] = 0
    
    if first_col_zero:
        for i in range(m):
            matrix[i][0] = 0

方法四：位运算优化

使用位运算记录需要置零的行和列。适用于矩阵规模较小的情况，但实际应用中较少使用。

def setZeroes(matrix):
    m, n = len(matrix), len(matrix[0])
    row_mask, col_mask = 0, 0
    
    for i in range(m):
        for j in range(n):
            if matrix[i][j] == 0:
                row_mask |= 1 << i
                col_mask |= 1 << j
    
    for i in range(m):
        for j in range(n):
            if (row_mask & (1 << i)) or (col_mask & (1 << j)):
                matrix[i][j] = 0

总结

矩阵置零问题可以通过多种方法解决，每种方法在时间复杂度和空间复杂度上各有优劣。实际应用中，方法二和方法三因其原地修改的特性更为常用。理解这些方法的差异有助于在实际问题中选择合适的解决方案。

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