【Leetcode】【python】Set Matrix Zeroes

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)

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])        for i in range(m):            if matrix[i][0] == 0:                first_col = True        for j in range(n):            if matrix[0][j] == 0:                first_row = True        for i in range(1, m):            for j in range(1, n):                if matrix[i][j] == 0:                    matrix[i][0] = matrix[0][j] = 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        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