【Leetcode】【python】Set Matrix Zeroes 矩阵置零

题目大意

如果矩阵中存在0，那么把0所在的行和列都置为0。要求在所给的矩阵上完成操作。
注意：最好的空间复杂度是常数空间

解题思路

参考：
https://www.hrwhisper.me/leetcode-set-matrix-zeroes/
https://shenjie1993.gitbooks.io/leetcode-python/073%20Set%20Matrix%20Zeroes.html

思路：
1.创建一个矩阵的拷贝，然后根据这个拷贝进行判断O(MN)
2.创建一个数组，记录矩阵为0的行和列下标O(m+n)
3.把有0的元素映射到首行和首列O(c)：详解看代码

代码

这里的复杂度是空间复杂度

O(c)

class Solution(object):
    def setZeroes(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: void Do not return anything, modify matrix in-place instead.
        """
        first_row = False
        first_col = False
        m = len(matrix)
        n = len(matrix[0])
        # 第一列若有0，first_col = True
        for i in range(m): 
            if matrix[i][0] == 0:
                first_col = True
        # 第一行若有0，first_row = True
        for j in range(n):
            if matrix[0][j] == 0:
                first_row = True
        # 若中间有0，同时映射到首行首列
        for i in range(1, m):
            for j in range(1, n):
                if matrix[i][j] == 0:
                    matrix[i][0] = matrix[0][j] = 0
        # 然后通过首行首列，将所有改为0的置为0
        for i in range(1, m):
            for j in range(1, n):
                if matrix[0][j] == 0 or matrix[i][0] == 0:
                    matrix[i][j] = 0
        # 最后再把保存好的首行首列是否改为0的信息拿出来置为0
        if first_row:
            for j in range(n):
                matrix[0][j] = 0
        if first_col:
            for i in range(m):
                matrix[i][0] = 0

O(m+n)

class Solution:
    # @param matrix, a list of lists of integers
    # RETURN NOTHING, MODIFY matrix IN PLACE.
    def setZeroes(self, matrix):
        m , n = len(matrix), len(matrix[0])
        row , col = [0 for i in range(m)] , [0 for i in range(n)]
        for i in range(m):
            for j in range(n):
                if not matrix[i][j]:
                    row[i]=col[j]=1
        for i in range(m):
            if row[i]:
                for j in range(n):
                    matrix[i][j]=0
 
        for j in range(n):
            if col[j]:
                for i in range(m):
                    matrix[i][j]=0

O(mn)

class Solution:
    # @param matrix, a list of lists of integers
    # RETURN NOTHING, MODIFY matrix IN PLACE.
    def setZeroes(self, matrix):
        m , n = len(matrix), len(matrix[0])
        temp = [[matrix[i][j] for j in range(n)] for i in range(m)]
        for i in range(m):
            for j in range(n):
                if not temp[i][j]:
                    self.setZero(i,j,n,m,matrix)
                    
    def setZero(self,row,col,n,m,matrix):
        for i in range(m):
            matrix[i][col]=0
        for j in range(n):
            matrix[row][j]=0

总结