刷题-力扣-面试题 01.07. 旋转矩阵(Brush questions – force buckle – interview question 01.07 Rotation matrix)

题目链接

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/rotate-matrix-lcci
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题目描述

给你一幅由 N × N 矩阵表示的图像,其中每个像素的大小为 4 字节。请你设计一种算法,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]
]

题目分析

  • 根据题目描述,将矩阵进行一次旋转
  • 模拟矩阵旋转,可发现矩阵分为四个部分

代码

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int row;
        int col;
        if (n % 2) {
            row = n / 2;
            col = n / 2 + 1;
        } else {
            row = n / 2;
            col = n / 2;
        }
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                int 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;
            }
        }
        return;
    }
};
————————

Title Link

Source: leetcode
Link: https://leetcode-cn.com/problems/rotate-matrix-lcci
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Title Description

Here’s a picture by n × N matrix represents the image, where the size of each pixel is 4 bytes. Please design an algorithm to rotate the image 90 degrees.

Can you do it without taking up additional memory space?

Example 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]
]

Example 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]
]

Topic analysis

  • Rotate the matrix once according to the title description
  • By simulating matrix rotation, it can be found that the matrix is divided into four parts

code

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n = matrix.size();
        int row;
        int col;
        if (n % 2) {
            row = n / 2;
            col = n / 2 + 1;
        } else {
            row = n / 2;
            col = n / 2;
        }
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                int 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;
            }
        }
        return;
    }
};