0048. Rotate Image

# 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]
]
``````

## 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
}

``````

Apr 8, 2023