leetcode 48. Rotate Image【旋转90】

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本文详细解析了LeetCode上的一道经典算法题——旋转图像。题目要求将一个n x n的二维矩阵顺时针旋转90度，且必须在原地修改矩阵，不能使用额外的矩阵进行操作。文章通过示例说明了如何实现这一目标，并提供了一段C++代码实现，帮助读者理解算法的具体步骤。

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https://leetcode.com/problems/rotate-image/description/

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]
]

自己画一画就出来了

要不是因为困应该不用写这么久的

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int n=matrix[0].size();
      //  printf("%d\n",n);
        for(int i=0;i<n/2;i++){
            for(int j=0;j<n-i*2-1;j++){
                int x=matrix[i][i+j];
               // printf("x=%d\n",x);
            //    printf("%d %d  ",i,i+j);printf("%d %d  ",n-i-1-j,i);printf("%d %d  ",n-i-1,n-1-i-j);printf("%d %d\n",i+j,n-i-1);
                matrix[i][i+j]=matrix[n-i-1-j][i];
                
                matrix[n-i-1-j][i]=matrix[n-i-1][n-i-j-1];
                
                matrix[n-i-1][n-i-j-1]=matrix[i+j][n-i-1];
                
                matrix[i+j][n-i-1]=x;
                
            }
        }        
    }
};