48. 旋转图像

English Version

题目描述

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。

 

示例 1：

输入：matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出：[[7,4,1],[8,5,2],[9,6,3]]

示例 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]]

 

提示：

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

 

解法

Python3

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        s, n = 0, len(matrix)
        while s < (n >> 1):
            e = n - s - 1
            for i in range(s, e):
                t = matrix[i][e]
                matrix[i][e] = matrix[s][i]
                matrix[s][i] = matrix[n - i - 1][s]
                matrix[n - i - 1][s] = matrix[e][n - i - 1]
                matrix[e][n - i - 1] = t
            s += 1

Java

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

C++

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {

        int n = matrix.size();
        if (n <= 1) return;

        // 先做转置
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                swap(matrix[i][j], matrix[j][i]);
            }
        }

        // 再做水平互换
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n / 2; j++) {
                swap(matrix[i][j], matrix[i][n - 1 - j]);
            }
        }
    }
};

JavaScript

/**
 * @param {number[][]} matrix
 * @return {void} Do not return anything, modify matrix in-place instead.
 */
var rotate = function (matrix) {
    const n = matrix.length;
    for (let i = 0; i < n; i++) {
        for (let j = 0; j <= i; j++) {
            [matrix[i][j], matrix[j][i]] = [matrix[j][i], matrix[i][j]];
        }
    }
    for (let i = 0, j = n - 1; i < j; i++, j--) {
        for (let k = 0; k < n; k++) {
            [matrix[k][i], matrix[k][j]] = [matrix[k][j], matrix[k][i]];
        }
    }
};

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

TypeScript

/**
 Do not return anything, modify matrix in-place instead.
 */
function rotate(matrix: number[][]): void {
    let n = matrix[0].length;
    for (let i = 0; i < Math.floor(n / 2); i++) {
        for (let j = 0; j < Math.floor((n + 1) / 2); j++) {
            let 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;
        }
    }
}

Rust

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

Go

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

	for i := 0; i < n; i++ {
		for j, k := 0, n-1; j < k; j, k = j+1, k-1 {
			matrix[i][j], matrix[i][k] = matrix[i][k], matrix[i][j]
		}
	}
}

...