Leetcode: 48.Rotate Image(Week4, Medium)-优快云博客

本文链接：https://blog.youkuaiyun.com/linwh8/article/details/78147799

本文介绍了一种高效的矩阵旋转方法，包括顺时针旋转和逆时针旋转的两种算法实现。顺时针旋转通过先翻转再转置的方式简化操作，而逆时针旋转则通过先倒置元素再转置矩阵来实现。

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本题采用了两种解法：常规解法与特殊解法

Leetcode 48
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]
]

题意：实现矩阵的顺时针旋转

思路：

I. 常规思路：总结规律。
- (0,0)–>(0,3)
- (0,1)–>(1,3)
- (0,2)–>(2,3)
- (0,3)–>(3,3)
- (1,0)–>(0,2)
- (1,1)–>(1,2)
- (1,2)–>(2,2)
- (1,3)–>(3,2)
- … (以此类推)
- 由以上我们能得到规律: (i, j) = (j, size()-i-1)
- 所以代码如下：

    /*
2017/10/1 Rotate_Image
思路：本题要求我们提供一个算法，将矩阵进行旋转。
      本题是顺时针旋转。可以通过找规律，来进行旋转！
 */
class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        vector<bool> isSwap(matrix.size()*matrix[0].size(), 0);
        for (int i = 0; i < matrix.size(); i++) {
            for (int j = 0; j < matrix[i].size(); j++) {
                if (!isSwap[i*matrix[i].size()+j]) {
                    int temp = matrix[j][matrix[i].size()-i-1];
                    matrix[j][matrix[i].size()-i-1] = matrix[i][j];
                    matrix[i][j] = temp;
                    isSwap[i*matrix[i].size()+j] = 1;
                    isSwap[j*matrix[i].size()+matrix[i].size()-i-1] = 1;
                }
            }
        }
    }
};

II. 特殊解法：先将行倒置，然后再对矩阵进行转置。

    /*
       clockwise rotate
       first reverse up to down, then swap the symmetry 
       1 2 3     7 8 9     7 4 1
       4 5 6  => 4 5 6  => 8 5 2
       7 8 9     1 2 3     9 6 3
    */
    void rotate(vector<vector<int> > &matrix) {
        reverse(matrix.begin(), matrix.end());
        for (int i = 0; i < matrix.size(); ++i) {
            for (int j = i + 1; j < matrix[i].size(); ++j)
                swap(matrix[i][j], matrix[j][i]);
        }
    }

III. 如果是逆时针呢，可以采用：先将各行的元素进行倒置，然后对矩阵转置

/*
  anticlockwise rotate
  first reverse left to right, then swap the symmetry
  1 2 3     3 2 1     3 6 9
  4 5 6  => 6 5 4  => 2 5 8
  7 8 9     9 8 7     1 4 7
*/
void anti_rotate(vector<vector<int> > &matrix) {
    for (auto vi : matrix) reverse(vi.begin(), vi.end());
    for (int i = 0; i < matrix.size(); ++i) {
        for (int j = i + 1; j < matrix[i].size(); ++j)
            swap(matrix[i][j], matrix[j][i]);
    }
}