Programmer interview golden classic-interview questions 01.07. Rotation matrix-find the law

1. Topic introduction

Give you an image represented by an N × N matrix, where each pixel is 4 bytes in size. Please design an algorithm to rotate the image 90 degrees.

Can it be done without taking up additional memory space?

 

Example 1:

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

Rotate the input matrix in place so that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]
Example 2:

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

Rotate the input matrix in place so that it becomes:
[
  [15,13, ​​2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11 ]
]

Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/rotate-matrix-lcci
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.

Two, problem-solving ideas

The following figure shows the change that occurs when the square rectangle is rotated by 90°: 4->1->2->3->4 (representing the position change during 90 clockwise rotation)

In a given matrix, divide the area in the upper left corner, and each element in the area can be rotated by 90° after three exchanges . Let the dimension of the matrix be n.

• swap(matrix[i][j], matrix[j][n-i])
• swap(matrix[i][j], matrix[n-i][n-j])
• swap(matrix[i][j], matrix[n-j][i])

Three, problem-solving code

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int len = matrix.size();
        int m = (len >> 1);   //短边，注意要加括号
        int n = (--len >> 1) + 1; //长边，考虑到N为奇数的情况，中心位置不发生变化
        for(int i = 0; i < m; ++i)
        {
            for(int j = 0; j < n; ++j)
            {
                swap(matrix[i][j], matrix[j][len-i]);
                swap(matrix[i][j], matrix[len-i][len-j]);
                swap(matrix[i][j], matrix[len-j][i]);
            }
        }
    }
};

Four, problem-solving results