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

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

### 题目描述

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

[
[7,4,1],
[8,5,2],
[9,6,3]
]
``````

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

[
[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;
}
};
``````
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Source: leetcode
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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]
],

[
[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]
],

[
[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;
}
};
``````