# [Swift]LeetCode832. 翻转图像 | Flipping an Image

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Given a binary matrix A, we want to flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed.  For example, flipping [1, 1, 0] horizontally results in [0, 1, 1].

To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0. For example, inverting [0, 1, 1] results in [1, 0, 0].

Example 1:

Input: [[1,1,0],[1,0,1],[0,0,0]]
Output: [[1,0,0],[0,1,0],[1,1,1]]
Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]].
Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]]


Example 2:

Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]
Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]
Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]].
Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]


Notes:

• 1 <= A.length = A[0].length <= 20
• 0 <= A[i][j] <= 1

输入: [[1,1,0],[1,0,1],[0,0,0]]

然后反转图片: [[1,0,0],[0,1,0],[1,1,1]]


输入: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]]

然后反转图片: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]]


• 1 <= A.length = A[0].length <= 20
• 0 <= A[i][j] <= 1

Runtime: 32 ms
Memory Usage: 18.7 MB
 1 class Solution {
2     func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] {
3         var a = Array<Array<Int>>()
4         for i in 0..<A.count{
5             var temp = Array<Int>()
6             for j in 0..<A.count{
7                 temp.append(1 - A[i][A.count - j - 1])
8             }
9             a.append(temp)
10         }
11         return a
12     }
13 }

32ms

1 class Solution {
2     func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] {
3         return A.map({$0.reversed().map({$0 == 1 ? 0 : 1})})
4     }
5 }

36ms

 1 class Solution {
2     func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] {
3         func inverse(_ int: Int) -> Int {
4             return int == 0 ? 1 : 0
5         }
6         return A.map { row in
7                       row.compactMap { element in
8                                inverse(element)
9                               }.reversed()
10         }
11     }
12 }

36ms

 1 class Solution {
2     func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] {
3         var result: [[Int]] = []
4         for row in A {
5             let temp = row.reversed().map{ $0 == 1 ? 0 : 1} 6 result.append(temp) 7 } 8 return result 9 } 10 } 36ms  1 class Solution { 2 func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] { 3 if A.count == 0 || A[0].count == 0{ 4 return A 5 } 6 7 var A = A 8 9 if A[0].count == 1 { 10 for x in 0..<A.count { 11 if A[x][0] == 1{ 12 A[x][0] = 0 13 }else{ 14 A[x][0] = 1 15 } 16 } 17 return A 18 } 19 20 let isOdd = A[0].count/2*2 != A[0].count 21 let checkYCount = isOdd ? (A[0].count/2 + 1) : (A[0].count/2) 22 for x in 0..<A.count { 23 for y in 0..<checkYCount { 24 if A[x][y] == A[x][A[0].count-y-1] { 25 if A[x][y] == 1{ 26 A[x][y] = 0 27 }else{ 28 A[x][y] = 1 29 } 30 A[x][A[0].count-y-1] = A[x][y] 31 } 32 } 33 } 34 return A 35 } 36 } 40ms  1 class Solution { 2 func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] { 3 if A == nil || A.count == 0 || A[0].count == 0 { 4 return A 5 } 6 let m = A.count, n = A[0].count 7 var res = [[Int]]() 8 for array in A { 9 res.append(array.reversed()) 10 } 11 12 for i in 0 ..< m { 13 for j in 0 ..< n { 14 if res[i][j] == 0 { 15 res[i][j] = 1 16 } else { 17 res[i][j] = 0 18 } 19 } 20 } 21 return res 22 } 23 } 52ms 1 class Solution { 2 func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] { 3 return A.map { 4$0.reversed().map { 1 - \$0 }
5         }
6     }
7 }

52ms

 1 class Solution {
2     func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] {
3         var countA = A.count
4         var res:[[Int]] = [[Int]](repeating:[Int](),count:countA)
5         for i in 0..<countA
6         {
7             for j in stride(from:countA - 1,through:0,by: -1)
8             {
9                 res[i].append(1 - A[i][j])
10             }
11         }
12         return res
13     }
14 }

76ms

 1 class Solution {
2     func flipAndInvertImage(_ A: [[Int]]) -> [[Int]] {
3         return A.map({ (nums) -> [Int] in
4             return flip(nums)
5         })
6     }
7
8     func flip(_ A: [Int]) -> [Int] {
9         return A.reversed().map { (num) -> Int in
10             return num == 0 ? 1 : 0
11         }
12     }
13 }

posted @ 2019-03-21 21:31 山青咏芝 阅读(...) 评论(...) 编辑 收藏