Rotate Image
描述
You are given an n × n
2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Follow up: Could you do this in-place?
分析
首先想到,纯模拟,从外到内一圈一圈的转,但这个方法太慢。
如下图,首先沿着副对角线翻转一次,然后沿着水平中线翻转一次。
或者,首先沿着水平中线翻转一次,然后沿着主对角线翻转一次。
代码 1
- Python
- Java
- C++
// Rotate Image
// 思路 1,时间复杂度O(n^2),空间复杂度O(1)
public class Solution {
public void rotate(final int[][] matrix) {
final int n = matrix.length;
for (int i = 0; i < n; ++i) // 沿着副对角线反转
for (int j = 0; j < n - i; ++j)
swap(matrix, i, j, n - 1 - j, n - 1 - i);
for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
for (int j = 0; j < n; ++j)
swap(matrix, i, j, n - 1 - i, j);
}
private static void swap(final int[][] matrix,
int i, int j, int p, int q) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[p][q];
matrix[p][q] = tmp;
}
};
// Rotate Image
// 思路 1,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
const int n = matrix.size();
for (int i = 0; i < n; ++i) // 沿着副对角线反转
for (int j = 0; j < n - i; ++j)
swap(matrix[i][j], matrix[n - 1 - j][n - 1 - i]);
for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
for (int j = 0; j < n; ++j)
swap(matrix[i][j], matrix[n - 1 - i][j]);
}
};
# Rotate Image
# 思路 1,时间复杂度O(n^2),空间复杂度O(1)
class Solution:
def rotate(self, matrix: list[list[int]]) -> None:
n = len(matrix)
# 沿着副对角线反转
for i in range(n):
for j in range(n - i):
self.swap(matrix, i, j, n - 1 - j, n - 1 - i)
# 沿着水平中线反转
for i in range(n // 2):
for j in range(n):
self.swap(matrix, i, j, n - 1 - i, j)
def swap(self, matrix: list[list[int]], i: int, j: int, p: int, q: int) -> None:
matrix[i][j], matrix[p][q] = matrix[p][q], matrix[i][j]
代码 2
- Java
- C++
// Rotate Image
// 思路 2,时间复杂度O(n^2),空间复杂度O(1)
public class Solution {
public void rotate(final int[][] matrix) {
final int n = matrix.length;
for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
for (int j = 0; j < n; ++j)
swap(matrix, i, j, n - 1 - i, j);
for (int i = 0; i < n; ++i) // 沿着主对角线反转
for (int j = i + 1; j < n; ++j)
swap(matrix, i, j, j, i);
}
private static void swap(final int[][] matrix,
int i, int j, int p, int q) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[p][q];
matrix[p][q] = tmp;
}
};
// Rotate Image
// 思路 2,时间复杂度O(n^2),空间复杂度O(1)
class Solution {
public:
void rotate(vector<vector<int>>& matrix) {
const int n = matrix.size();
for (int i = 0; i < n / 2; ++i) // 沿着水平中线反转
for (int j = 0; j < n; ++j)
swap(matrix[i][j], matrix[n - 1 - i][j]);
for (int i = 0; i < n; ++i) // 沿着主对角线反转
for (int j = i + 1; j < n; ++j)
swap(matrix[i][j], matrix[j][i]);
}
};
# Rotate Image
# 思路 2,时间复杂度O(n^2),空间复杂度O(1)
class Solution:
def rotate(self, matrix: list[list[int]]) -> None:
n = len(matrix)
# 沿着水平中线反转
for i in range(n // 2):
for j in range(n):
self.swap(matrix, i, j, n - 1 - i, j)
# 沿着主对角线反转
for i in range(n):
for j in range(i + 1, n):
self.swap(matrix, i, j, j, i)
def swap(self, matrix: list[list[int]], i: int, j: int, p: int, q: int) -> None:
matrix[i][j], matrix[p][q] = matrix[p][q], matrix[i][j]