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feat: update lc problems #2166

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Dec 29, 2023
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feat: update lc problems #2166

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30 changes: 29 additions & 1 deletion solution/0000-0099/0052.N-Queens II/README.md
View file
Original file line number Diff line number Diff line change
@@ -66,7 +66,7 @@
```python
class Solution:
def totalNQueens(self, n: int) -> int:
def dfs(i):
def dfs(i: int):
if i == n:
nonlocal ans
ans += 1
@@ -184,6 +184,34 @@ func totalNQueens(n int) (ans int) {
}
```

### **TypeScript**

```ts
function totalNQueens(n: number): number {
const cols: boolean[] = Array(10).fill(false);
const dg: boolean[] = Array(20).fill(false);
const udg: boolean[] = Array(20).fill(false);
let ans = 0;
const dfs = (i: number) => {
if (i === n) {
++ans;
return;
}
for (let j = 0; j < n; ++j) {
let [a, b] = [i + j, i - j + n];
if (cols[j] || dg[a] || udg[b]) {
continue;
}
cols[j] = dg[a] = udg[b] = true;
dfs(i + 1);
cols[j] = dg[a] = udg[b] = false;
}
};
dfs(0);
return ans;
}
```

### **...**

```
30 changes: 29 additions & 1 deletion solution/0000-0099/0052.N-Queens II/README_EN.md
View file
Original file line number Diff line number Diff line change
@@ -54,7 +54,7 @@ The time complexity is $O(n!)$, and the space complexity is $O(n)$. Here, $n$ is
```python
class Solution:
def totalNQueens(self, n: int) -> int:
def dfs(i):
def dfs(i: int):
if i == n:
nonlocal ans
ans += 1
@@ -170,6 +170,34 @@ func totalNQueens(n int) (ans int) {
}
```

### **TypeScript**

```ts
function totalNQueens(n: number): number {
const cols: boolean[] = Array(10).fill(false);
const dg: boolean[] = Array(20).fill(false);
const udg: boolean[] = Array(20).fill(false);
let ans = 0;
const dfs = (i: number) => {
if (i === n) {
++ans;
return;
}
for (let j = 0; j < n; ++j) {
let [a, b] = [i + j, i - j + n];
if (cols[j] || dg[a] || udg[b]) {
continue;
}
cols[j] = dg[a] = udg[b] = true;
dfs(i + 1);
cols[j] = dg[a] = udg[b] = false;
}
};
dfs(0);
return ans;
}
```

### **...**

```
42 changes: 21 additions & 21 deletions solution/0000-0099/0052.N-Queens II/Solution.py
View file
Original file line number Diff line number Diff line change
@@ -1,21 +1,21 @@
class Solution:
def totalNQueens(self, n: int) -> int:
def dfs(i):
if i == n:
nonlocal ans
ans += 1
return
for j in range(n):
a, b = i + j, i - j + n
if cols[j] or dg[a] or udg[b]:
continue
cols[j] = dg[a] = udg[b] = True
dfs(i + 1)
cols[j] = dg[a] = udg[b] = False

cols = [False] * 10
dg = [False] * 20
udg = [False] * 20
ans = 0
dfs(0)
return ans
class Solution:
def totalNQueens(self, n: int) -> int:
def dfs(i: int):
if i == n:
nonlocal ans
ans += 1
return
for j in range(n):
a, b = i + j, i - j + n
if cols[j] or dg[a] or udg[b]:
continue
cols[j] = dg[a] = udg[b] = True
dfs(i + 1)
cols[j] = dg[a] = udg[b] = False
cols = [False] * 10
dg = [False] * 20
udg = [False] * 20
ans = 0
dfs(0)
return ans
23 changes: 23 additions & 0 deletions solution/0000-0099/0052.N-Queens II/Solution.ts
View file
Original file line number Diff line number Diff line change
@@ -0,0 +1,23 @@
function totalNQueens(n: number): number {
const cols: boolean[] = Array(10).fill(false);
const dg: boolean[] = Array(20).fill(false);
const udg: boolean[] = Array(20).fill(false);
let ans = 0;
const dfs = (i: number) => {
if (i === n) {
++ans;
return;
}
for (let j = 0; j < n; ++j) {
let [a, b] = [i + j, i - j + n];
if (cols[j] || dg[a] || udg[b]) {
continue;
}
cols[j] = dg[a] = udg[b] = true;
dfs(i + 1);
cols[j] = dg[a] = udg[b] = false;
}
};
dfs(0);
return ans;
}
2 changes: 1 addition & 1 deletion solution/0000-0099/0063.Unique Paths II/README.md
View file
Original file line number Diff line number Diff line change
@@ -55,7 +55,7 @@

首先初始化 $dp$ 第一列和第一行的所有值，然后遍历其它行和列，有两种情况：

- 若 $obstacleGrid[i][j] = 1$，说明路径数为 $0$，那么 $dp[i][j] = 0 ￥；
- 若 $obstacleGrid[i][j] = 1$，说明路径数为 $0$，那么 $dp[i][j] = 0$；
- 若 ￥ obstacleGrid[i][j] = 0$，则 $dp[i][j] = dp[i - 1][j] + dp[i][j - 1]$。

最后返回 $dp[m - 1][n - 1]$ 即可。
2 changes: 1 addition & 1 deletion solution/0000-0099/0065.Valid Number/README.md
View file
Original file line number Diff line number Diff line change
@@ -91,7 +91,7 @@

遍历完字符串后，返回 `true`。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串长度。
时间复杂度 $O(n)$，其中 $n$ 为字符串长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

117 changes: 55 additions & 62 deletions solution/0000-0099/0069.Sqrt(x)/README.md
View file
Original file line number Diff line number Diff line change
@@ -41,7 +41,15 @@

<!-- 这里可写通用的实现逻辑 -->

二分查找。
**方法一：二分查找**

我们定义二分查找的左边界 $l = 0$，右边界 $r = x$，然后在 $[l, r]$ 范围内查找平方根。

在每一步查找中，我们找出中间值 $mid = (l + r + 1) / 2$，如果 $mid > x / mid$，说明平方根在 $[l, mid - 1]$ 范围内，我们令 $r = mid - 1$；否则说明平方根在 $[mid, r]$ 范围内，我们令 $l = mid$。

查找结束后，返回 $l$ 即可。

时间复杂度 $O(\log x)$，空间复杂度 $O(1)$。

<!-- tabs:start -->

@@ -52,15 +60,14 @@
```python
class Solution:
def mySqrt(self, x: int) -> int:
left, right = 0, x
while left < right:
mid = (left + right + 1) >> 1
# mid*mid <= x
if mid <= x // mid:
left = mid
l, r = 0, x
while l < r:
mid = (l + r + 1) >> 1
if mid > x // mid:
r = mid - 1
else:
right = mid - 1
return left
l = mid
return l
```

### **Java**
@@ -70,17 +77,16 @@ class Solution:
```java
class Solution {
public int mySqrt(int x) {
int left = 0, right = x;
while (left < right) {
int mid = (left + right + 1) >>> 1;
if (mid <= x / mid) {
// mid*mid <= x
left = mid;
int l = 0, r = x;
while (l < r) {
int mid = (l + r + 1) >>> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
right = mid - 1;
l = mid;
}
}
return left;
return l;
}
}
```
@@ -91,15 +97,16 @@ class Solution {
class Solution {
public:
int mySqrt(int x) {
long long left = 0, right = x;
while (left < right) {
long long mid = left + ((right - left + 1) >> 1);
if (mid <= x / mid)
left = mid;
else
right = mid - 1;
int l = 0, r = x;
while (l < r) {
int mid = (l + r + 1ll) >> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
l = mid;
}
}
return (int) left;
return l;
}
};
```
@@ -108,16 +115,7 @@ public:

```go
func mySqrt(x int) int {
left, right := 0, x
for left < right {
mid := left + (right-left+1)>>1
if mid <= x/mid {
left = mid
} else {
right = mid - 1
}
}
return left
return sort.Search(x+1, func(i int) bool { return i*i > x }) - 1
}
```

@@ -129,17 +127,16 @@ func mySqrt(x int) int {
* @return {number}
*/
var mySqrt = function (x) {
let left = 0;
let right = x;
while (left < right) {
const mid = (left + right + 1) >>> 1;
if (mid <= x / mid) {
left = mid;
let [l, r] = [0, x];
while (l < r) {
const mid = (l + r + 1) >> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
right = mid - 1;
l = mid;
}
}
return left;
return l;
};
```

@@ -148,20 +145,16 @@ var mySqrt = function (x) {
```cs
public class Solution {
public int MySqrt(int x) {
int left = 0, right = x;
while (left < right)
{
int mid = left + right + 1 >> 1;
if (mid <= x / mid)
{
left = mid;
}
else
{
right = mid - 1;
int l = 0, r = x;
while (l < r) {
int mid = (l + r + 1) >>> 1;
if (mid > x / mid) {
r = mid - 1;
} else {
l = mid;
}
}
return left;
return l;
}
}
```
@@ -171,19 +164,19 @@ public class Solution {
```rust
impl Solution {
pub fn my_sqrt(x: i32) -> i32 {
if x < 2 {
return x;
}
let mut l = 1;
let mut r = x / 2;
let mut l = 0;
let mut r = x;

while l < r {
let mid = (l + r + 1) >> 1;
if x / mid < mid {
let mid = (l + r + 1) / 2;

if mid > x / mid {
r = mid - 1;
} else {
l = mid;
}
}

l
}
}
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