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Aug 2, 2023
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feat: add solutions to lc problem: No.1914 #1372

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261 changes: 203 additions & 58 deletions solution/1900-1999/1914.Cyclically Rotating a Grid/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -49,6 +49,14 @@

<!-- 这里可写通用的实现逻辑 -->

**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。

<!-- tabs:start -->

### **Python3**
Expand All @@ -58,41 +66,37 @@
```python
class Solution:
def rotateGrid(self, grid: List[List[int]], k: int) -> List[List[int]]:
def rotate(grid, s1, e1, s2, e2, k):
t = []
for j in range(e2, e1, -1):
t.append(grid[s1][j])
for i in range(s1, s2):
t.append(grid[i][e1])
for j in range(e1, e2):
t.append(grid[s2][j])
for i in range(s2, s1, -1):
t.append(grid[i][e2])
k %= len(t)
t = t[-k:] + t[:-k]
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(e2, e1, -1):
grid[s1][j] = t[k]
for j in range(p, n - p - 1):
grid[p][j] = nums[k]
k += 1
for i in range(s1, s2):
grid[i][e1] = t[k]
for i in range(p, m - p - 1):
grid[i][n - p - 1] = nums[k]
k += 1
for j in range(e1, e2):
grid[s2][j] = t[k]
for j in range(n - p - 1, p, -1):
grid[m - p - 1][j] = nums[k]
k += 1
for i in range(s2, s1, -1):
grid[i][e2] = t[k]
for i in range(m - p - 1, p, -1):
grid[i][p] = nums[k]
k += 1

m, n = len(grid), len(grid[0])
s1 = e1 = 0
s2, e2 = m - 1, n - 1
while s1 <= s2 and e1 <= e2:
rotate(grid, s1, e1, s2, e2, k)
s1 += 1
e1 += 1
s2 -= 1
e2 -= 1
for p in range(min(m, n) >> 1):
rotate(p, k)
return grid
```

Expand All @@ -102,53 +106,194 @@ class Solution:

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

public int[][] rotateGrid(int[][] grid, int k) {
int m = grid.length, n = grid[0].length;
int s1 = 0, e1 = 0;
int s2 = m - 1, e2 = n - 1;
while (s1 <= s2 && e1 <= e2) {
rotate(grid, s1++, e1++, s2--, e2--, 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[][] grid, int s1, int e1, int s2, int e2, int k) {
List<Integer> t = new ArrayList<>();
for (int j = e2; j > e1; --j) {
t.add(grid[s1][j]);
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 = s1; i < s2; ++i) {
t.add(grid[i][e1]);
for (int i = p; i < m - p - 1; ++i) {
nums.add(grid[i][n - p - 1]);
}
for (int j = e1; j < e2; ++j) {
t.add(grid[s2][j]);
for (int j = n - p - 1; j > p; --j) {
nums.add(grid[m - p - 1][j]);
}
for (int i = s2; i > s1; --i) {
t.add(grid[i][e2]);
for (int i = m - p - 1; i > p; --i) {
nums.add(grid[i][p]);
}
int n = t.size();
k %= n;
int l = nums.size();
k %= l;
if (k == 0) {
return;
}
k = n - k;
for (int j = e2; j > e1; --j) {
grid[s1][j] = t.get(k);
k = (k + 1) % n;
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);
}
}
}
```

### **C++**

```cpp
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;
}
};
```

### **Go**

```go
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
}
```

### **TypeScript**

```ts
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 (int i = s1; i < s2; ++i) {
grid[i][e1] = t.get(k);
k = (k + 1) % n;
for (let i = p; i < m - p - 1; ++i) {
grid[i][n - p - 1] = nums[k++ % l];
}
for (int j = e1; j < e2; ++j) {
grid[s2][j] = t.get(k);
k = (k + 1) % n;
for (let j = n - p - 1; j > p; --j) {
grid[m - p - 1][j] = nums[k++ % l];
}
for (int i = s2; i > s1; --i) {
grid[i][e2] = t.get(k);
k = (k + 1) % n;
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;
}
```

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