48. Rotate Image-优快云博客

本文介绍了一个高效的矩阵旋转算法，该算法能在不使用额外空间的情况下将矩阵顺时针旋转90度。通过逆序操作和元素交换实现了原地旋转，提供了详细的实现步骤及代码示例。

48. Rotate Image

题目

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

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

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

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

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

解析

本题带有技巧，如果允许申请空间，可以直接按列取，放入到新空间的一行即可

class Solution_48 {
public:
    void rotate(vector<vector<int>>& matrix) {

        // 总的位置坐标关系：rotate[j][n-1-i]=a[i][j]
        int n = matrix.size();
        reverse(matrix.begin(), matrix.end()); //a[n-1-i][j]=a[i][j] 
        for (int i = 0; i < n;i++)
        {
            for (int j = i + 1; j < n;j++)
            {
                swap(matrix[i][j],matrix[j][i]); // a[i][j]=a[j][i]
            }
        }
        return;
    }
};