[Swift]LeetCode73. 矩阵置零 | 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],  [1,1]
]
Output: 
[
  [1,  [0,0],1]
]

Example 2:

Input: 
[
  [0,2,  [3,4,5,2],3,5]
]
Output: 
[
  [0,0]
]

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]
]
输出: 
[
  [1,1]
]

示例 2:

输入: 
[
  [0,5]
]
输出: 
[
  [0,0]
]

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,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,1)
 9                     helper(&matrix,2)
10                     helper(&matrix,3)
11                     helper(&matrix,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,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 }

[Swift]LeetCode73. 矩阵置零 | Set Matrix Zeroes

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