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] ]
Follow up:
- A straight forward solution using O(mn) space is probably a bad idea.
- A simple improvement uses O(m + n) space, but still not the best solution.
- Could you devise a constant space solution?
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。请使用原地算法。
示例 1:
输入: [ [1,1,1], [1,0,1], [1,1,1] ] 输出: [ [1,0,1], [0,0,0], [1,0,1] ]
示例 2:
输入: [ [0,1,2,0], [3,4,5,2], [1,3,1,5] ] 输出: [ [0,0,0,0], [0,4,5,0], [0,3,1,0] ]
进阶:
- 一个直接的解决方案是使用 O(mn) 的额外空间，但这并不是一个好的解决方案。
- 一个简单的改进方案是使用 O(m + n) 的额外空间，但这仍然不是最好的解决方案。
- 你能想出一个常数空间的解决方案吗？
40ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3 var rows = [Int]()
4 var cols = [Int]()
5
6 for i in 0 ..< matrix.count {
7 for j in 0 ..< matrix[i].count {
8 if matrix[i][j] == 0 {
9 rows.append(i)
10 cols.append(j)
11 }
12 }
13 }
14 // Set rows to 0
15 for col in cols {
16 for i in 0 ..< matrix.count {
17 matrix[i][col] = 0
18 }
19 }
20 // Set cols to 0
21 for row in rows {
22 for j in 0 ..< matrix[row].count {
23 matrix[row][j] = 0
24 }
25 }
26 }
27 }
44ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3 for i in 0..<matrix.count {
4 for j in 0..<matrix[i].count {
5 if matrix[i][j] == 0 {
6 matrix[i][j] = -9999
7 }
8 }
9 }
10
11 for i in 0..<matrix.count {
12 for j in 0..<matrix[i].count {
13 if matrix[i][j] == -9999 {
14 for c in 0..<matrix[i].count {
15 if (matrix[i][c] != -9999) {
16 matrix[i][c] = 0
17 }
18 }
19 for r in 0..<matrix.count {
20 if (matrix[r][j] != -9999) {
21 matrix[r][j] = 0
22 }
23 }
24 matrix[i][j] = 0
25 }
26 }
27 }
28 }
29 }
44ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3 var rows = [Int]()
4 var cols = [Int]()
5 for i in 0..<matrix.count {
6 for j in 0..<matrix[i].count {
7 if matrix[i][j] == 0 {
8 rows.append(i)
9 cols.append(j)
10 }
11 }
12 }
13 for row in rows {
14 matrix[row] = Array(repeating: 0, count: matrix[row].count)
15 }
16 for col in cols {
17 for i in 0..<matrix.count {
18 matrix[i][col] = 0
19 }
20 }
21 }
22 }
48ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3 if matrix.count == 0 {
4 return
5 }
6
7 //查找第一行是否有0
8 var row0HasZore = false
9 for value in matrix[0] {
10 if value == 0 {
11 row0HasZore = true
12 break
13 }
14 }
15
16 if matrix.count > 1 {
17 //查找每一行
18 for i in 1..<matrix.count {
19 var rowiHasZore = false
20 //查找该行是否有0，并置第一行该列数为0
21 for j in 0..<matrix[0].count {
22 if matrix[i][j] == 0 {
23 rowiHasZore = true
24 matrix[0][j] = 0
25 }
26 }
27 //如果该行有0，就置该行所有数为0
28 if rowiHasZore {
29 matrix[i].replaceSubrange(0..<matrix[0].count, with: [Int](repeating: 0, count: matrix[0].count))
30 }
31 }
32 //查找第一行是否有0，有则把整列赋值0
33 for j in 0..<matrix[0].count {
34 if matrix[0][j] == 0 {
35 for i in 0..<matrix.count {
36 matrix[i][j] = 0
37 }
38 }
39 }
40 }
41 //如果第一行有0，则把该行赋值为0
42 if row0HasZore {
43 matrix[0].replaceSubrange(0..<matrix[0].count, with: [Int](repeating: 0, count: matrix[0].count))
44 }
45 }
46 }
76ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3 var rows = [Bool](repeating: false, count: matrix.count)
4 var cols = [Bool](repeating: false, count: matrix[0].count)
5 for (i, row) in matrix.enumerated() {
6 for (j, num) in row.enumerated() {
7 if num == 0 {
8 rows[i] = true
9 cols[j] = true
10 }
11 }
12 }
13 for (i, row) in matrix.enumerated() {
14 for (j, _) in row.enumerated() {
15 if rows[i] || cols[j] {
16 matrix[i][j] = 0
17 }
18 }
19 }
20 }
21 }
160ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3 var isCol:Bool = false
4 var R:Int = matrix.count
5 var C:Int = matrix[0].count
6 for i in 0..<R
7 {
8 //因为第一行和第一列的第一个单元是相同的，即矩阵[0][0]
9 //我们可以为第一行/列使用一个附加变量。
10 //对于这个解决方案，我们使用第一列的附加变量。
11 //使用第一行的矩阵[0][0]
12 if matrix[i][0] == 0 {isCol = true}
13 for j in 1..<C
14 {
15 //如果元素为零，则将相应行和列的第一个元素设置为0
16 if matrix[i][j] == 0
17 {
18 matrix[0][j] = 0
19 matrix[i][0] = 0
20 }
21 }
22 }
23 //再次迭代数组并使用第一行和第一列，更新元素
24 for i in 1..<R
25 {
26 for j in 1..<C
27 {
28 if matrix[i][0] == 0 || matrix[0][j] == 0
29 {
30 matrix[i][j] = 0
31 }
32 }
33 }
34 //是否需要将第一行设置为0
35 if matrix[0][0] == 0
36 {
37 for j in 0..<C
38 {
39 matrix[0][j] = 0
40 }
41 }
42 //是否需要将第一列设置为0
43 if isCol
44 {
45 for i in 0..<R
46 {
47 matrix[i][0] = 0
48 }
49 }
50 }
51 }
236ms
1 class Solution {
2 func setZeroes(_ matrix: inout [[Int]]) {
3
4 for i in matrix.indices {
5 for j in matrix[i].indices {
6 //print(i,j, matrix[i][j])
7 if matrix[i][j] == 0 {
8 helper(&matrix, i, j, 1)
9 helper(&matrix, i, j, 2)
10 helper(&matrix, i, j, 3)
11 helper(&matrix, i, j, 4)
12 }
13 }
14 }
15
16 //print(matrix)
17 for i in matrix.indices {
18 for j in matrix[i].indices {
19 if matrix[i][j] == Int.max {
20 matrix[i][j] = 0
21 }
22 }
23 }
24 }
25
26 func helper(_ matrix: inout[[Int]], _ i: Int, _ j: Int, _ dir: Int) {
27 //print(i, j, dir)
28 var i = i
29 var j = j
30 if dir == 1 {
31 while i >= 0 {
32 if (matrix[i][j] != 0) {
33 matrix[i][j] = Int.max
34 }
35
36 i -= 1
37 }
38 } else if dir == 2 {
39 while i < matrix.count {
40 if (matrix[i][j] != 0) {
41 matrix[i][j] = Int.max
42 }
43 i += 1
44 }
45 } else if dir == 3 {
46 while j >= 0 {
47 if (matrix[i][j] != 0) {
48 matrix[i][j] = Int.max
49 }
50 j -= 1
51 }
52 } else {
53 while j < matrix[i].count {
54 if (matrix[i][j] != 0) {
55 matrix[i][j] = Int.max
56 }
57 j += 1
58 }
59 }
60 }
61 }
原文地址：https://www.cnblogs.com/strengthen/p/9928348.html
时间： 2024-10-10 00:32:26