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48. Rotate Image

Description

You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise).

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:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [[7,4,1],[8,5,2],[9,6,3]]

Example 2:

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

 

Constraints:

  • n == matrix.length == matrix[i].length
  • 1 <= n <= 20
  • -1000 <= matrix[i][j] <= 1000

Solutions

Solution 1: In-place Rotation

Thinking

The first idea is a new matrix with \(\textit{matrix}[j][n-1-i] \gets \textit{matrix}[i][j]\). Correct, \(O(n^2)\) time and space. \(n \le 20\) fits, but the problem requires in-place.

The extra matrix is the bottleneck. A \(90^\circ\) clockwise turn factors into two in-place flips: reverse upside-down, then transpose across the main diagonal.

\((i, j)\) goes to \((n-1-i, j)\), then to \((j, n-1-i)\) — the target. Two rounds of swaps, \(O(1)\) extra space.

According to the problem requirements, we need to rotate \(\text{matrix}[i][j]\) to \(\text{matrix}[j][n - i - 1]\).

We can first flip the matrix upside down, i.e., swap \(\text{matrix}[i][j]\) with \(\text{matrix}[n - i - 1][j]\), and then flip the matrix along the main diagonal, i.e., swap \(\text{matrix}[i][j]\) with \(\text{matrix}[j][i]\). This way, we can rotate \(\text{matrix}[i][j]\) to \(\text{matrix}[j][n - i - 1]\).

The time complexity is \(O(n^2)\), where \(n\) is the side length of the matrix. The space complexity is \(O(1)\).

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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]

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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;
            }
        }
    }
}

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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]);
            }
        }
    }
};

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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]
        }
    }
}

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/**
 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;
        }
    }
}

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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;
            }
        }
    }
}

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/**
 * @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]];
        }
    }
};

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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;
            }
        }
    }
}

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