1 """ 2 Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place. 3 Example 1: 4 Input: 5 [ 6 [1,1,1], 7 [1,0,1], 8 [1,1,1] 9 ] 10 Output: 11 [ 12 [1,0,1], 13 [0,0,0], 14 [1,0,1] 15 ] 16 Example 2: 17 Input: 18 [ 19 [0,1,2,0], 20 [3,4,5,2], 21 [1,3,1,5] 22 ] 23 Output: 24 [ 25 [0,0,0,0], 26 [0,4,5,0], 27 [0,3,1,0] 28 ] 29 """ 30 """ 31 自己的方法：一遍AC 32 空间复杂度为O(m*n),需要进一步优化 33 """ 34 class Solution: 35 def setZeroes(self, matrix): 36 """ 37 Do not return anything, modify matrix in-place instead. 38 """ 39 queue = set() 40 for i in range(len(matrix)): 41 for j in range(len(matrix[0])): 42 if matrix[i][j] == 0: 43 queue.add((i, j)) 44 for x, y in queue: 45 for n in range(len(matrix[x])): 46 matrix[x][n] = 0 47 for m in range(len(matrix)): 48 matrix[m][y] = 0 49 """ 50 解法二：自己优化了一下 51 """ 52 class Solution2: 53 def setZeroes(self, matrix): 54 """ 55 Do not return anything, modify matrix in-place instead. 56 """ 57 row = set() 58 col = set() 59 for i in range(len(matrix)): 60 for j in range(len(matrix[0])): 61 if matrix[i][j] == 0: 62 row.add(i) 63 col.add(j) 64 for x in row: 65 matrix[x] = [0]*len(matrix[x]) 66 for y in col: 67 for i in range(len(matrix)): 68 matrix[i][y] = 0 69 """ 70 解法三：空间复杂度最低 71 将需要改变的值先变为'a' 72 最后再遍历一遍将值为'a'的变为0 73 """ 74 class Solution3: 75 def setZeroes(self, matrix): 76 """ 77 Do not return anything, modify matrix in-place instead. 78 """ 79 row = len(matrix) 80 col = len(matrix[0]) 81 for i in range(row): 82 for j in range(col): 83 if matrix[i][j] == 0: 84 for temp in range(row): 85 if matrix[temp][j] != 0: 86 matrix[temp][j] = 'a' 87 for temp in range(col): 88 if matrix[i][temp] != 0: 89 matrix[i][temp] = 'a' 90 for i in range(row): 91 for j in range(col): 92 if matrix[i][j] == 'a': 93 matrix[i][j] = 0