旋转图像

Math is crucial In some cases。

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

我的（ 20 ms ）


func rotate(_ matrix: inout [[Int]]) {  
    var sub:Int = 0
    let len=matrix.count
    for i in 0..<len {
        for j in i+1..<len {
            sub=matrix[i][j]
            matrix[i][j]=matrix[j][i]
            matrix[j][i]=sub
        }
        matrix[i].reverse()
    }    
}

先对数组进行转置，然后倒叙行就完成了，注意第二个循环
for j in i+1..<len

16 ms

func rotateImage1(_ matrix: inout [[Int]]) {
        let n = matrix.count
        for i in 0..<n/2 {
            for j in i..<(n - 1 - i) {
                let tmp = matrix[i][j]
                matrix[i][j] = matrix[n - 1 - j][i]
                matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]
                matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]
                matrix[j][n - 1 - i] = tmp
            }
        }
    }

对于当前位置，计算旋转后的新位置，然后再计算下一个新位置，第四个位置又变成当前位置了，所以这个方法每次循环换四个数字。

1 2 3      7 2 1      7 4 1
4 5 6 ->   4 5 6  ->  8 5 2
7 8 9      9 8 3      9 6 3

还有一种

func rotate(_ matrix: inout [[Int]]) {
        let n = matrix.count
        for i in 0..<(n - 1) {
            for j in 0..<(n - i) {
                let temp = matrix[i][j]
                matrix[i][j] = matrix[n - 1 - j][n - 1 - i]
                matrix[n - j - 1][n - i - 1] = temp
            }
        }
        
        for i in 0..<n/2 {
            for j in 0..<n {
                let temp = matrix[i][j]
                matrix[i][j] = matrix[n - 1 - i][j]
                matrix[n - 1 - i][j] = temp
            }
        }
    }