leetcode73 Set Matrix Zeroes

"""
Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
Example 1:
Input:
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output:
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]
Example 2:
Input:
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output:
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]
"""
"""
自己的方法：一遍AC
空间复杂度为O(m*n),需要进一步优化
"""
class Solution:
    def setZeroes(self, matrix):
        """
        Do not return anything, modify matrix in-place instead.
        """
        queue = set()
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j] == 0:
                    queue.add((i, j))
        for x, y in queue:
            for n in range(len(matrix[x])):
                matrix[x][n] = 0
            for m in range(len(matrix)):
                matrix[m][y] = 0
"""
解法二：自己优化了一下
"""
class Solution2:
    def setZeroes(self, matrix):
        """
        Do not return anything, modify matrix in-place instead.
        """
        row = set()
        col = set()
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j] == 0:
                    row.add(i)
                    col.add(j)
        for x in row:
            matrix[x] = [0]*len(matrix[x])
        for y in col:
            for i in range(len(matrix)):
                matrix[i][y] = 0
"""
解法三：空间复杂度最低
将需要改变的值先变为'a'
最后再遍历一遍将值为'a'的变为0
"""
class Solution3:
    def setZeroes(self, matrix):
        """
        Do not return anything, modify matrix in-place instead.
        """
        row = len(matrix)
        col = len(matrix[0])
        for i in range(row):
            for j in range(col):
                if matrix[i][j] == 0:
                    for temp in range(row):
                        if matrix[temp][j] != 0:
                            matrix[temp][j] = 'a'
                    for temp in range(col):
                        if matrix[i][temp] != 0:
                            matrix[i][temp] = 'a'
        for i in range(row):
            for j in range(col):
                if matrix[i][j] == 'a':
                    matrix[i][j] = 0