README.md

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

原地旋转，i 的范围是 [0, n/2)，j 的范围是 [i, n-1-i)。

Python3

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        n = len(matrix)
        for i in range(n // 2):
            for j in range(i, n - 1 - i):
                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

Java

class Solution {
    public void rotate(int[][] matrix) {
        int n = matrix.length;
        for (int i = 0; i < n / 2; ++i) {
            for (int j = i; j < n - 1 - i; ++j) {
                int 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;
            }
        }
    }
}

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 / 2; i++) {
        for (let j = i; j < n - i - 1; j++) {
            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)
	r, c := n/2, (n+1)/2
	for i := 0; i < r; i++ {
		for j := 0; j < c; j++ {
			temp := 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] = temp
		}
	}
}

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