Leetcode: 48.Rotate Image(Week4, Medium)

本题采用了两种解法：常规解法与特殊解法

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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]);    }}

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以上内容皆为本人观点，欢迎大家匹配和指导，我们一起探讨！

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