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feat: update lc problems #3778

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Nov 19, 2024
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feat: update lc problems #3778

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8 changes: 4 additions & 4 deletions lcp/LCP 22. 黑白方格画/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -203,7 +203,7 @@ class Solution {
if k == 0 || k == n * n {
return 1
}

func combination(_ n: Int, _ r: Int) -> Int {
guard r <= n else { return 0 }
if r == 0 || r == n { return 1 }
Expand All @@ -213,9 +213,9 @@ class Solution {
}
return result
}

var ans = 0

for i in 0...n {
for j in 0...n {
let paintedCells = n * (i + j) - i * j
Expand All @@ -224,7 +224,7 @@ class Solution {
}
}
}

return ans
}
}
Expand Down
6 changes: 3 additions & 3 deletions lcp/LCP 24. 数字游戏/README.md
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Expand Up @@ -320,12 +320,12 @@ class MedianFinder {
init() {}

func addNum(_ num: Int) {

upperHeap.append(num)
sumUpper += num
upperHeap.sort()


if let minUpper = upperHeap.first {
upperHeap.removeFirst()
lowerHeap.append(minUpper)
Expand All @@ -334,7 +334,7 @@ class MedianFinder {
lowerHeap.sort(by: >)
}


if lowerHeap.count > upperHeap.count + 1 {
if let maxLower = lowerHeap.first {
lowerHeap.removeFirst()
Expand Down
8 changes: 4 additions & 4 deletions lcp/LCP 25. 古董键盘/README.md
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Expand Up @@ -203,26 +203,26 @@ class Solution {
for i in 0...n {
c[i][0] = 1
}

for i in 1...n {
for j in 1...k {
c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod
}
}

var f = Array(repeating: Array(repeating: 0, count: 27), count: n + 1)
for j in 0..<27 {
f[0][j] = 1
}

for i in 1...n {
for j in 1..<27 {
for h in 0...min(i, k) {
f[i][j] = (f[i][j] + (f[i - h][j - 1] * c[i][h]) % mod) % mod
}
}
}

return f[n][26]
}
}
Expand Down
36 changes: 30 additions & 6 deletions solution/0800-0899/0867.Transpose Matrix/README.md
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Expand Up @@ -58,7 +58,15 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：模拟

我们记矩阵 $\textit{matrix}$ 的行数为 $m$，列数为 $n$。根据转置的定义，转置后的矩阵 $\textit{ans}$ 的行数为 $n$，列数为 $m$。

对于 $\textit{ans}$ 中的任意位置 $(i,j)$，其对应于矩阵 $\textit{matrix}$ 中的位置 $(j,i)$。因此，我们遍历矩阵 $\textit{matrix}$ 中的每个元素，将其转置到 $\textit{ans}$ 中相应的位置。

遍历结束后，返回 $\textit{ans}$ 即可。

时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵 $\textit{matrix}$ 的行数和列数。忽略答案的空间消耗，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -95,9 +103,11 @@ public:
vector<vector<int>> transpose(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> ans(n, vector<int>(m));
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
};
Expand All @@ -119,6 +129,21 @@ func transpose(matrix [][]int) [][]int {
}
```

#### TypeScript

```ts
function transpose(matrix: number[][]): number[][] {
const [m, n] = [matrix.length, matrix[0].length];
const ans: number[][] = Array.from({ length: n }, () => Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
```

#### JavaScript

```js
Expand All @@ -127,9 +152,8 @@ func transpose(matrix [][]int) [][]int {
* @return {number[][]}
*/
var transpose = function (matrix) {
const m = matrix.length;
const n = matrix[0].length;
const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
const [m, n] = [matrix.length, matrix[0].length];
const ans = Array.from({ length: n }, () => Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
Expand Down
36 changes: 30 additions & 6 deletions solution/0800-0899/0867.Transpose Matrix/README_EN.md
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Expand Up @@ -56,7 +56,15 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Simulation

Let $m$ be the number of rows and $n$ be the number of columns in the matrix $\textit{matrix}$. According to the definition of transpose, the transposed matrix $\textit{ans}$ will have $n$ rows and $m$ columns.

For any position $(i, j)$ in $\textit{ans}$, it corresponds to the position $(j, i)$ in the matrix $\textit{matrix}$. Therefore, we traverse each element in the matrix $\textit{matrix}$ and transpose it to the corresponding position in $\textit{ans}$.

After the traversal, we return $\textit{ans}$.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the matrix $\textit{matrix}$, respectively. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -93,9 +101,11 @@ public:
vector<vector<int>> transpose(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> ans(n, vector<int>(m));
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
};
Expand All @@ -117,6 +127,21 @@ func transpose(matrix [][]int) [][]int {
}
```

#### TypeScript

```ts
function transpose(matrix: number[][]): number[][] {
const [m, n] = [matrix.length, matrix[0].length];
const ans: number[][] = Array.from({ length: n }, () => Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
```

#### JavaScript

```js
Expand All @@ -125,9 +150,8 @@ func transpose(matrix [][]int) [][]int {
* @return {number[][]}
*/
var transpose = function (matrix) {
const m = matrix.length;
const n = matrix[0].length;
const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
const [m, n] = [matrix.length, matrix[0].length];
const ans = Array.from({ length: n }, () => Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
Expand Down
8 changes: 5 additions & 3 deletions solution/0800-0899/0867.Transpose Matrix/Solution.cpp
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Expand Up @@ -3,9 +3,11 @@ class Solution {
vector<vector<int>> transpose(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
vector<vector<int>> ans(n, vector<int>(m));
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j)
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
};
};
5 changes: 2 additions & 3 deletions solution/0800-0899/0867.Transpose Matrix/Solution.js
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Expand Up @@ -3,9 +3,8 @@
* @return {number[][]}
*/
var transpose = function (matrix) {
const m = matrix.length;
const n = matrix[0].length;
const ans = new Array(n).fill(0).map(() => new Array(m).fill(0));
const [m, n] = [matrix.length, matrix[0].length];
const ans = Array.from({ length: n }, () => Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
Expand Down
10 changes: 10 additions & 0 deletions solution/0800-0899/0867.Transpose Matrix/Solution.ts
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Original file line number Diff line number Diff line change
@@ -0,0 +1,10 @@
function transpose(matrix: number[][]): number[][] {
const [m, n] = [matrix.length, matrix[0].length];
const ans: number[][] = Array.from({ length: n }, () => Array(m).fill(0));
for (let i = 0; i < n; ++i) {
for (let j = 0; j < m; ++j) {
ans[i][j] = matrix[j][i];
}
}
return ans;
}
84 changes: 41 additions & 43 deletions solution/0800-0899/0868.Binary Gap/README.md
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Expand Up @@ -68,7 +68,11 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：位运算

我们用两个指针 $\textit{pre}$ 和 $\textit{cur}$ 分别表示上一个和当前的 $1$ 的位置，初始时 $\textit{pre} = 100$, $\textit{cur} = 0$，然后遍历 $n$ 的二进制表示，当遇到 $1$ 时，计算当前位置和上一个 $1$ 的位置之间的距离，并更新答案。

时间复杂度 $O(\log n)$，其中 $n$ 是题目给定的整数。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -77,12 +81,13 @@ tags:
```python
class Solution:
def binaryGap(self, n: int) -> int:
ans, j = 0, -1
for i in range(32):
ans = 0
pre, cur = inf, 0
while n:
if n & 1:
if j != -1:
ans = max(ans, i - j)
j = i
ans = max(ans, cur - pre)
pre = cur
cur += 1
n >>= 1
return ans
```
Expand All @@ -93,13 +98,12 @@ class Solution:
class Solution {
public int binaryGap(int n) {
int ans = 0;
for (int i = 0, j = -1; n != 0; ++i, n >>= 1) {
if ((n & 1) == 1) {
if (j != -1) {
ans = Math.max(ans, i - j);
}
j = i;
for (int pre = 100, cur = 0; n != 0; n >>= 1) {
if (n % 2 == 1) {
ans = Math.max(ans, cur - pre);
pre = cur;
}
++cur;
}
return ans;
}
Expand All @@ -113,11 +117,12 @@ class Solution {
public:
int binaryGap(int n) {
int ans = 0;
for (int i = 0, j = -1; n; ++i, n >>= 1) {
for (int pre = 100, cur = 0; n != 0; n >>= 1) {
if (n & 1) {
if (j != -1) ans = max(ans, i - j);
j = i;
ans = max(ans, cur - pre);
pre = cur;
}
++cur;
}
return ans;
}
Expand All @@ -127,36 +132,31 @@ public:
#### Go

```go
func binaryGap(n int) int {
ans := 0
for i, j := 0, -1; n != 0; i, n = i+1, n>>1 {
if (n & 1) == 1 {
if j != -1 && ans < i-j {
ans = i - j
}
j = i
func binaryGap(n int) (ans int) {
for pre, cur := 100, 0; n != 0; n >>= 1 {
if n&1 == 1 {
ans = max(ans, cur-pre)
pre = cur
}
cur++
}
return ans
return
}
```

#### TypeScript

```ts
function binaryGap(n: number): number {
let res = 0;
let j = -1;
for (let i = 0; n !== 0; i++) {
let ans = 0;
for (let pre = 100, cur = 0; n; n >>= 1) {
if (n & 1) {
if (j !== -1) {
res = Math.max(res, i - j);
}
j = i;
ans = Math.max(ans, cur - pre);
pre = cur;
}
n >>= 1;
++cur;
}
return res;
return ans;
}
```

Expand All @@ -165,20 +165,18 @@ function binaryGap(n: number): number {
```rust
impl Solution {
pub fn binary_gap(mut n: i32) -> i32 {
let mut res = 0;
let mut i = 0;
let mut j = -1;
let mut ans = 0;
let mut pre = 100;
let mut cur = 0;
while n != 0 {
if (n & 1) == 1 {
if j != -1 {
res = res.max(i - j);
}
j = i;
if n % 2 == 1 {
ans = ans.max(cur - pre);
pre = cur;
}
cur += 1;
n >>= 1;
i += 1;
}
res
ans
}
}
```
Expand Down
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