Question

面试题 01.07. 旋转矩阵

English Version

题目描述

给定一幅由N × N矩阵表示的图像，其中每个像素的大小为4字节，编写一种方法，将图像旋转90度。

不占用额外内存空间能否做到？

 

示例 1:

给定 matrix =
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

原地旋转输入矩阵，使其变为:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

示例 2:

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

原地旋转输入矩阵，使其变为:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

解法

方法一：原地翻转

根据题目要求，我们实际上需要将 $matrix[i][j]$ 旋转至 $matrix[j][n - i - 1]$。

我们可以先对矩阵进行上下翻转，即 $matrix[i][j]$ 和 $matrix[n - i - 1][j]$ 进行交换，然后再对矩阵进行主对角线翻转，即 $matrix[i][j]$ 和 $matrix[j][i]$ 进行交换。这样就能将 $matrix[i][j]$ 旋转至 $matrix[j][n - i - 1]$ 了。

时间复杂度 $O(n^2)$，其中 $n$ 是矩阵的边长。空间复杂度 $O(1)$。

class Solution:
  def rotate(self, matrix: List[List[int]]) -> None:
    n = len(matrix)
    for i in range(n >> 1):
      for j in range(n):
        matrix[i][j], matrix[n - i - 1][j] = matrix[n - i - 1][j], matrix[i][j]
    for i in range(n):
      for j in range(i):
        matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]

class Solution {
  public void rotate(int[][] matrix) {
    int n = matrix.length;
    for (int i = 0; i < n >> 1; ++i) {
      for (int j = 0; j < n; ++j) {
        int t = matrix[i][j];
        matrix[i][j] = matrix[n - i - 1][j];
        matrix[n - i - 1][j] = t;
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < i; ++j) {
        int t = matrix[i][j];
        matrix[i][j] = matrix[j][i];
        matrix[j][i] = t;
      }
    }
  }
}

class Solution {
public:
  void rotate(vector<vector<int>>& matrix) {
    int n = matrix.size();
    for (int i = 0; i < n >> 1; ++i) {
      for (int j = 0; j < n; ++j) {
        swap(matrix[i][j], matrix[n - i - 1][j]);
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < i; ++j) {
        swap(matrix[i][j], matrix[j][i]);
      }
    }
  }
};

func rotate(matrix [][]int) {
  n := len(matrix)
  for i := 0; i < n>>1; i++ {
    for j := 0; j < n; j++ {
      matrix[i][j], matrix[n-i-1][j] = matrix[n-i-1][j], matrix[i][j]
    }
  }
  for i := 0; i < n; i++ {
    for j := 0; j < i; j++ {
      matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]
    }
  }
}

/**
 Do not return anything, modify matrix in-place instead.
 */
function rotate(matrix: number[][]): void {
  matrix.reverse();
  for (let i = 0; i < matrix.length; ++i) {
    for (let j = 0; j < i; ++j) {
      const t = matrix[i][j];
      matrix[i][j] = matrix[j][i];
      matrix[j][i] = t;
    }
  }
}

impl Solution {
  pub fn rotate(matrix: &mut Vec<Vec<i32>>) {
    let n = matrix.len();
    for i in 0..n / 2 {
      for j in 0..n {
        let t = matrix[i][j];
        matrix[i][j] = matrix[n - i - 1][j];
        matrix[n - i - 1][j] = t;
      }
    }
    for i in 0..n {
      for j in 0..i {
        let t = matrix[i][j];
        matrix[i][j] = matrix[j][i];
        matrix[j][i] = t;
      }
    }
  }
}

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var rotate = function (matrix) {
  matrix.reverse();
  for (let i = 0; i < matrix.length; ++i) {
    for (let j = 0; j < i; ++j) {
      [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
    }
  }
};

public class Solution {
  public void Rotate(int[][] matrix) {
    int n = matrix.Length;
    for (int i = 0; i < n >> 1; ++i) {
      for (int j = 0; j < n; ++j) {
        int t = matrix[i][j];
        matrix[i][j] = matrix[n - i - 1][j];
        matrix[n - i - 1][j] = t;
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < i; ++j) {
        int t = matrix[i][j];
        matrix[i][j] = matrix[j][i];
        matrix[j][i] = t;
      }
    }
  }
}