题目描述

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]
]

思路

题目不让分配空间，希望以巧妙的方式解题。这类题目虽然明知道有巧妙的解法，思维却很容易被数字搅乱或被不重要的细节干扰，需要静下心对规律进行梳理和抽象。

算法1 按圈旋转

想象矩阵转动的过程，容易产生依次旋转覆盖的想法。如果一个数一个数的旋转，就只需要保存一个值。从外到内将n*n的矩阵看成$(n+1)/2$个圈，对于每个圈，只需将最上面一排的每个数字作为开启旋转的钥匙，依次覆盖右下左的每个数字就可以了。

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        int len=matrix.size()-1;
        for(int i=0;i<(len+1)/2;i++)//圈数
        {
            for(int k=i;k<len-i;k++)
            {
                int tmp=matrix[k][len-i];
                matrix[k][len-i]=matrix[i][k];
                matrix[i][k]=matrix[len-k][i];
                matrix[len-k][i]=matrix[len-i][len-k];
                matrix[len-i][len-k]=tmp;
            }                        
        }
    }
};

算法2 翻转矩阵

如果对规律再进行抽象，还可以发现先沿对角翻转再沿中线翻转正好是旋转90°的位置，这个代码就简单很多，但是很难想到了。

内心永远是彩虹色！