1.问题描述
给定一个 n × n 的二维矩阵 matrix 表示一个图像。请你将图像顺时针旋转 90 度。
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。
2.测试用例
用例1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出:[[7,4,1],[8,5,2],[9,6,3]]
用例2
输入:matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
输出:[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
用例3:
输入:matrix = [[1]]
输出:[[1]]
用例4
输入:matrix = [[1,2],[3,4]]
输出:[[3,1],[4,2]]
3.提示
matrix.length == n
matrix[i].length == n
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000
4.图解-代码
1.数学方法:矩阵转置 + 反转
图解
code
public void rotate(int[][] matrix) {
matrixPrint(matrix);
if (matrix.length < 2) {
matrixPrint(matrix);
return;
}
for (int i = 0; i < matrix.length; i++) {
for (int j = i; j < matrix.length; j++) {
exchangeMatrix(matrix, i, j);
}
reverse(matrix[i]);
}
matrixPrint(matrix);
}
public void matrixPrint(int[][] matrix) {
for (int[] ints : matrix) {
for (int anInt : ints) {
System.out.print(anInt + "\t");
}
System.out.println();
}
}
public void reverse(int[] nums) {
int start = 0;
int end = nums.length - 1;
while (start < end) {
exchange(nums, start, end);
start++;
end--;
}
}
public void exchangeMatrix(int[][] nums, int i, int j) {
int tmp = nums[i][j];
nums[i][j] = nums[j][i];
nums[j][i] = tmp;
}
分治-矩阵旋转
图解
![image-20211116190656615](/Users/xq/Library/Application Support/typora-user-images/image-20211116190656615.png)
code
public void rotate(int[][] matrix) {
int n = matrix.length;
matrixPrint(matrix);
if (n < 2) {
matrixPrint(matrix);
return;
}
for (int i = 0; i < (n + 1) / 2; i++) {
for (int j = 0; j < n / 2; j++) {
int 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;
}
}
matrixPrint(matrix);
}
public void matrixPrint(int[][] matrix) {
for (int[] ints : matrix) {
for (int anInt : ints) {
System.out.print(anInt + "\t");
}
System.out.println();
}
}