48. Rotate Image

题目

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 1:

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]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

题目大意

给定一个 n × n 的二维矩阵表示一个图像。将图像顺时针旋转 90 度。说明：你必须在原地旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要使用另一个矩阵来旋转图像。

解题思路

• 给出一个二维数组，要求顺时针旋转 90 度。
• 这一题比较简单，按照题意做就可以。这里给出 2 种旋转方法的实现，顺时针旋转和逆时针旋转。

    /*
     * clockwise rotate 顺时针旋转
     * first reverse up to down, then swap the symmetry 
     * 1 2 3     7 8 9     7 4 1
     * 4 5 6  => 4 5 6  => 8 5 2
     * 7 8 9     1 2 3     9 6 3
    */
    void rotate(vector<vector<int> > &matrix) {
        reverse(matrix.begin(), matrix.end());
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = i + 1; j < matrix[i].size(); ++j)
                swap(matrix[i][j], matrix[j][i]);
        }
    }
    
    /*
     * anticlockwise rotate 逆时针旋转
     * first reverse left to right, then swap the symmetry
     * 1 2 3     3 2 1     3 6 9
     * 4 5 6  => 6 5 4  => 2 5 8
     * 7 8 9     9 8 7     1 4 7
    */
    void anti_rotate(vector<vector<int> > &matrix) {
        for (auto vi : matrix) reverse(vi.begin(), vi.end());
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = i + 1; j < matrix[i].size(); ++j)
                swap(matrix[i][j], matrix[j][i]);
        }
    }
    

代码

package leetcode

// 解法一
func rotate(matrix [][]int) {
	length := len(matrix)
	// rotate by diagonal 对角线变换
	for i := 0; i < length; i++ {
		for j := i + 1; j < length; j++ {
			matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
		}
	}
	// rotate by vertical centerline 竖直轴对称翻转
	for i := 0; i < length; i++ {
		for j := 0; j < length/2; j++ {
			matrix[i][j], matrix[i][length-j-1] = matrix[i][length-j-1], matrix[i][j]
		}
	}
}

// 解法二
func rotate1(matrix [][]int) {
	n := len(matrix)
	if n == 1 {
		return
	}
	/* rotate clock-wise = 1. transpose matrix => 2. reverse(matrix[i])

	1   2  3  4      1   5  9  13        13  9  5  1
	5   6  7  8  =>  2   6  10 14  =>    14  10 6  2
	9  10 11 12      3   7  11 15        15  11 7  3
	13 14 15 16      4   8  12 16        16  12 8  4

	*/

	for i := 0; i < n; i++ {
		// transpose, i=rows, j=columns
		// j = i+1, coz diagonal elements didn't change in a square matrix
		for j := i + 1; j < n; j++ {
			swap(matrix, i, j)
		}
		// reverse each row of the image
		matrix[i] = reverse(matrix[i])
	}
}

// swap changes original slice's i,j position
func swap(nums [][]int, i, j int) {
	nums[i][j], nums[j][i] = nums[j][i], nums[i][j]
}

// reverses a row of image, matrix[i]
func reverse(nums []int) []int {
	var lp, rp = 0, len(nums) - 1

	for lp < rp {
		nums[lp], nums[rp] = nums[rp], nums[lp]
		lp++
		rp--
	}
	return nums
}

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