Question

1914. 循环轮转矩阵

English Version

题目描述

给你一个大小为 m x n 的整数矩阵 grid​​​ ，其中 m 和 n 都是 偶数 ；另给你一个整数 k 。

矩阵由若干层组成，如下图所示，每种颜色代表一层：

矩阵的循环轮转是通过分别循环轮转矩阵中的每一层完成的。在对某一层进行一次循环旋转操作时，层中的每一个元素将会取代其 逆时针 方向的相邻元素。轮转示例如下：

返回执行 k 次循环轮转操作后的矩阵。

 

示例 1：

输入：grid = [[40,10],[30,20]], k = 1
输出：[[10,20],[40,30]]
解释：上图展示了矩阵在执行循环轮转操作时每一步的状态。

示例 2：

输入：grid = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]], k = 2
输出：[[3,4,8,12],[2,11,10,16],[1,7,6,15],[5,9,13,14]]
解释：上图展示了矩阵在执行循环轮转操作时每一步的状态。

 

提示：

• m == grid.length
• n == grid[i].length
• 2 <= m, n <= 50
• m 和 n 都是 偶数
• 1 <= grid[i][j] <= 5000
• 1 <= k <= 109

解法

方法一：逐层模拟

我们先计算得到矩阵的层数 $p$，然后从外到内逐层模拟循环轮转的过程。

对于每一层，我们按照顺时针方向，将上、右、下、左四条边的元素依次放入数组 $nums$ 中。记数组 $nums$ 的长度为 $l$。接下来，我们将 $k$ 模 $l$。然后从数组的第 $k$ 个位置开始，将数组中的元素依次放回矩阵的上、右、下、左四条边。

时间复杂度 $O(m \times n)$，空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。

class Solution:
  def rotateGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
    def rotate(p: int, k: int):
      nums = []
      for j in range(p, n - p - 1):
        nums.append(grid[p][j])
      for i in range(p, m - p - 1):
        nums.append(grid[i][n - p - 1])
      for j in range(n - p - 1, p, -1):
        nums.append(grid[m - p - 1][j])
      for i in range(m - p - 1, p, -1):
        nums.append(grid[i][p])
      k %= len(nums)
      if k == 0:
        return
      nums = nums[k:] + nums[:k]
      k = 0
      for j in range(p, n - p - 1):
        grid[p][j] = nums[k]
        k += 1
      for i in range(p, m - p - 1):
        grid[i][n - p - 1] = nums[k]
        k += 1
      for j in range(n - p - 1, p, -1):
        grid[m - p - 1][j] = nums[k]
        k += 1
      for i in range(m - p - 1, p, -1):
        grid[i][p] = nums[k]
        k += 1

    m, n = len(grid), len(grid[0])
    for p in range(min(m, n) >> 1):
      rotate(p, k)
    return grid

class Solution {
  private int m;
  private int n;
  private int[][] grid;

  public int[][] rotateGrid(int[][] grid, int k) {
    m = grid.length;
    n = grid[0].length;
    this.grid = grid;
    for (int p = 0; p < Math.min(m, n) / 2; ++p) {
      rotate(p, k);
    }
    return grid;
  }

  private void rotate(int p, int k) {
    List<Integer> nums = new ArrayList<>();
    for (int j = p; j < n - p - 1; ++j) {
      nums.add(grid[p][j]);
    }
    for (int i = p; i < m - p - 1; ++i) {
      nums.add(grid[i][n - p - 1]);
    }
    for (int j = n - p - 1; j > p; --j) {
      nums.add(grid[m - p - 1][j]);
    }
    for (int i = m - p - 1; i > p; --i) {
      nums.add(grid[i][p]);
    }
    int l = nums.size();
    k %= l;
    if (k == 0) {
      return;
    }
    for (int j = p; j < n - p - 1; ++j) {
      grid[p][j] = nums.get(k++ % l);
    }
    for (int i = p; i < m - p - 1; ++i) {
      grid[i][n - p - 1] = nums.get(k++ % l);
    }
    for (int j = n - p - 1; j > p; --j) {
      grid[m - p - 1][j] = nums.get(k++ % l);
    }
    for (int i = m - p - 1; i > p; --i) {
      grid[i][p] = nums.get(k++ % l);
    }
  }
}

class Solution {
public:
  vector<vector<int>> rotateGrid(vector<vector<int>>& grid, int k) {
    int m = grid.size(), n = grid[0].size();
    auto rotate = [&](int p, int k) {
      vector<int> nums;
      for (int j = p; j < n - p - 1; ++j) {
        nums.push_back(grid[p][j]);
      }
      for (int i = p; i < m - p - 1; ++i) {
        nums.push_back(grid[i][n - p - 1]);
      }
      for (int j = n - p - 1; j > p; --j) {
        nums.push_back(grid[m - p - 1][j]);
      }
      for (int i = m - p - 1; i > p; --i) {
        nums.push_back(grid[i][p]);
      }
      int l = nums.size();
      k %= l;
      if (k == 0) {
        return;
      }
      for (int j = p; j < n - p - 1; ++j) {
        grid[p][j] = nums[k++ % l];
      }
      for (int i = p; i < m - p - 1; ++i) {
        grid[i][n - p - 1] = nums[k++ % l];
      }
      for (int j = n - p - 1; j > p; --j) {
        grid[m - p - 1][j] = nums[k++ % l];
      }
      for (int i = m - p - 1; i > p; --i) {
        grid[i][p] = nums[k++ % l];
      }
    };
    for (int p = 0; p < min(m, n) / 2; ++p) {
      rotate(p, k);
    }
    return grid;
  }
};

func rotateGrid(grid [][]int, k int) [][]int {
  m, n := len(grid), len(grid[0])

  rotate := func(p, k int) {
    nums := []int{}
    for j := p; j < n-p-1; j++ {
      nums = append(nums, grid[p][j])
    }
    for i := p; i < m-p-1; i++ {
      nums = append(nums, grid[i][n-p-1])
    }
    for j := n - p - 1; j > p; j-- {
      nums = append(nums, grid[m-p-1][j])
    }
    for i := m - p - 1; i > p; i-- {
      nums = append(nums, grid[i][p])
    }
    l := len(nums)
    k %= l
    if k == 0 {
      return
    }
    for j := p; j < n-p-1; j++ {
      grid[p][j] = nums[k]
      k = (k + 1) % l
    }
    for i := p; i < m-p-1; i++ {
      grid[i][n-p-1] = nums[k]
      k = (k + 1) % l
    }
    for j := n - p - 1; j > p; j-- {
      grid[m-p-1][j] = nums[k]
      k = (k + 1) % l
    }
    for i := m - p - 1; i > p; i-- {
      grid[i][p] = nums[k]
      k = (k + 1) % l
    }
  }

  for i := 0; i < m/2 && i < n/2; i++ {
    rotate(i, k)
  }
  return grid
}

function rotateGrid(grid: number[][], k: number): number[][] {
  const m = grid.length;
  const n = grid[0].length;
  const rotate = (p: number, k: number) => {
    const nums: number[] = [];
    for (let j = p; j < n - p - 1; ++j) {
      nums.push(grid[p][j]);
    }
    for (let i = p; i < m - p - 1; ++i) {
      nums.push(grid[i][n - p - 1]);
    }
    for (let j = n - p - 1; j > p; --j) {
      nums.push(grid[m - p - 1][j]);
    }
    for (let i = m - p - 1; i > p; --i) {
      nums.push(grid[i][p]);
    }
    const l = nums.length;
    k %= l;
    if (k === 0) {
      return;
    }
    for (let j = p; j < n - p - 1; ++j) {
      grid[p][j] = nums[k++ % l];
    }
    for (let i = p; i < m - p - 1; ++i) {
      grid[i][n - p - 1] = nums[k++ % l];
    }
    for (let j = n - p - 1; j > p; --j) {
      grid[m - p - 1][j] = nums[k++ % l];
    }
    for (let i = m - p - 1; i > p; --i) {
      grid[i][p] = nums[k++ % l];
    }
  };
  for (let p = 0; p < Math.min(m, n) >> 1; ++p) {
    rotate(p, k);
  }
  return grid;
}