https://leetcode.com/problems/rotate-image/
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]
]
其中row为行数,col为列数
- 起始位置:
matrix[row][col]
- 逆时针90度位置:
matrix[n - col - 1][row]
- 逆时针180度位置:
matrix[n- row - 1][n- col - 1]
- 逆时针270度位置:
matrix[col][n - row - 1]
matrix[j][n-1 - i] = matrix[i][j]
class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
// transpose matrix
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
int tmp = matrix[j][i];
matrix[j][i] = matrix[i][j];
matrix[i][j] = tmp;
}
}
// reverse each row
for (int i = 0; i < n; i++) {
for (int j = 0; j < n / 2; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[i][n - j - 1];
matrix[j][n - j - 1] = tmp;
}
}
}
}
public class Solution {
public void rotate(int[][] matrix) {
// 计算圈数
int n = matrix.length, lvl = n / 2;
for(int i = 0; i < lvl; i++){
for(int j = i; j < n - i - 1; j++){
// 左上和左下交换
swap(matrix, i, j, j, n - i - 1);
// 左上和右下交换
swap(matrix, i, j, n - i - 1, n - j - 1);
// 左上和右上交换
swap(matrix, i, j, n - j - 1, i);
}
}
}
private void swap(int[][] matrix, int i1, int j1, int i2, int j2){
int tmp = matrix[i1][j1];
matrix[i1][j1] = matrix[i2][j2];
matrix[i2][j2] = tmp;
}
}
Approach #2 Rotate four rectangles
class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < n / 2 + n % 2; i++) {
for (int j = 0; j < n / 2; j++) {
int[] tmp = new int[4];
int row = i;
int col = j;
for (int k = 0; k < 4; k++) {
tmp[k] = matrix[row][col];
int x = row;
row = col;
col = n - 1 - x;
}
for (int k = 0; k < 4; k++) {
matrix[row][col] = tmp[(k + 3) % 4];
int x = row;
row = col;
col = n - 1 - x;
}
}
}
}
}
Approach #3 Rotate four rectangels in one single loop
class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < (n + 1) / 2; i ++) {
for (int j = 0; j < n / 2; j++) {
int temp = matrix[n - 1 - j][i];
matrix[n - 1 - j][i] = matrix[n - 1 - i][n - j - 1];
matrix[n - 1 - i][n - j - 1] = matrix[j][n - 1 -i];
matrix[j][n - 1 - i] = matrix[i][j];
matrix[i][j] = temp;
}
}
}
}