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Nov 21, 2024
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[pull] main from doocs:main #208

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088bf63
feat: add solutions to lc problem: No.370 (#3796)
yanglbme Nov 21, 2024
f1b58d0
docs: update lcof problem: No.046 (#3795)
lingxier Nov 21, 2024
83d6171
feat: update solutions to lc problem: No.0496 (#3797)
yanglbme Nov 21, 2024
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6 changes: 3 additions & 3 deletions lcof/面试题46. 把数字翻译成字符串/README.md
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Expand Up @@ -40,12 +40,12 @@ edit_url: https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9

我们先将数字 `num` 转为字符串 $s$，字符串 $s$ 的长度记为 $n$。

然后我们设计一个函数 $dfs(i)$，表示从第 $i$ 个数字开始的不同翻译的数目。那么答案就是 $dfs(0)$。
然后我们设计一个函数 $dfs(i)$，表示从索引为 $i$ 的数字开始的不同翻译的数目。那么答案就是 $dfs(0)$。

函数 $dfs(i)$ 的计算如下：

- 如果 $i \ge n - 1$，说明已经翻译到最后一个数字，只有一种翻译方法，返回 $1$；
- 否则，我们可以选择翻译第 $i$ 个数字，此时翻译方法数目为 $dfs(i + 1)$；如果第 $i$ 个数字和第 $i + 1$ 个数字可以组成一个有效的字符（即 $s[i] == 1$ 或者 $s[i] == 2$ 且 $s[i + 1] \lt 6$），那么我们还可以选择翻译第 $i$ 和第 $i + 1$ 个数字，此时翻译方法数目为 $dfs(i + 2)$。因此 $dfs(i) = dfs(i+1) + dfs(i+2)$。
- 如果 $i \ge n - 1$，说明已经翻译到最后一个数字或者越过最后一个字符，均只有一种翻译方法，返回 $1$；
- 否则，我们可以选择翻译索引为 $i$ 的数字，此时翻译方法数目为 $dfs(i + 1)$；如果索引为 $i$ 的数字和索引为 $i + 1$ 的数字可以组成一个有效的字符（即 $s[i] == 1$ 或者 $s[i] == 2$ 且 $s[i + 1] \lt 6$），那么我们还可以选择翻译索引为 $i$ 和索引为 $i + 1$ 的数字，此时翻译方法数目为 $dfs(i + 2)$。因此 $dfs(i) = dfs(i+1) + dfs(i+2)$。

过程中我们可以使用记忆化搜索，将已经计算过的 $dfs(i)$ 的值存储起来，避免重复计算。

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241 changes: 161 additions & 80 deletions solution/0300-0399/0370.Range Addition/README.md
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Expand Up @@ -101,12 +101,16 @@ class Solution {
public:
vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
vector<int> d(length);
for (auto& e : updates) {
for (const auto& e : updates) {
int l = e[0], r = e[1], c = e[2];
d[l] += c;
if (r + 1 < length) d[r + 1] -= c;
if (r + 1 < length) {
d[r + 1] -= c;
}
}
for (int i = 1; i < length; ++i) {
d[i] += d[i - 1];
}
for (int i = 1; i < length; ++i) d[i] += d[i - 1];
return d;
}
};
Expand All @@ -131,6 +135,24 @@ func getModifiedArray(length int, updates [][]int) []int {
}
```

#### TypeScript

```ts
function getModifiedArray(length: number, updates: number[][]): number[] {
const d: number[] = Array(length).fill(0);
for (const [l, r, c] of updates) {
d[l] += c;
if (r + 1 < length) {
d[r + 1] -= c;
}
}
for (let i = 1; i < length; ++i) {
d[i] += d[i - 1];
}
return d;
}
```

#### JavaScript

```js
Expand All @@ -140,7 +162,7 @@ func getModifiedArray(length int, updates [][]int) []int {
* @return {number[]}
*/
var getModifiedArray = function (length, updates) {
const d = new Array(length).fill(0);
const d = Array(length).fill(0);
for (const [l, r, c] of updates) {
d[l] += c;
if (r + 1 < length) {
Expand Down Expand Up @@ -177,82 +199,74 @@ var getModifiedArray = function (length, updates) {

```python
class BinaryIndexedTree:
def __init__(self, n):
__slots__ = "n", "c"

def __init__(self, n: int):
self.n = n
self.c = [0] * (n + 1)

@staticmethod
def lowbit(x):
return x & -x

def update(self, x, delta):
def update(self, x: int, delta: int) -> None:
while x <= self.n:
self.c[x] += delta
x += BinaryIndexedTree.lowbit(x)
x += x & -x

def query(self, x):
def query(self, x: int) -> int:
s = 0
while x:
s += self.c[x]
x -= BinaryIndexedTree.lowbit(x)
x -= x & -x
return s


class Solution:
def getModifiedArray(self, length: int, updates: List[List[int]]) -> List[int]:
tree = BinaryIndexedTree(length)
for start, end, inc in updates:
tree.update(start + 1, inc)
tree.update(end + 2, -inc)
for l, r, c in updates:
tree.update(l + 1, c)
tree.update(r + 2, -c)
return [tree.query(i + 1) for i in range(length)]
```

#### Java

```java
class Solution {
public int[] getModifiedArray(int length, int[][] updates) {
BinaryIndexedTree tree = new BinaryIndexedTree(length);
for (int[] e : updates) {
int start = e[0], end = e[1], inc = e[2];
tree.update(start + 1, inc);
tree.update(end + 2, -inc);
}
int[] ans = new int[length];
for (int i = 0; i < length; ++i) {
ans[i] = tree.query(i + 1);
}
return ans;
}
}

class BinaryIndexedTree {
private int n;
private int[] c;

public BinaryIndexedTree(int n) {
this.n = n;
c = new int[n + 1];
this.c = new int[n + 1];
}

public void update(int x, int delta) {
while (x <= n) {
for (; x <= n; x += x & -x) {
c[x] += delta;
x += lowbit(x);
}
}

public int query(int x) {
int s = 0;
while (x > 0) {
for (; x > 0; x -= x & -x) {
s += c[x];
x -= lowbit(x);
}
return s;
}
}

public static int lowbit(int x) {
return x & -x;
class Solution {
public int[] getModifiedArray(int length, int[][] updates) {
var tree = new BinaryIndexedTree(length);
for (var e : updates) {
int l = e[0], r = e[1], c = e[2];
tree.update(l + 1, c);
tree.update(r + 2, -c);
}
int[] ans = new int[length];
for (int i = 0; i < length; ++i) {
ans[i] = tree.query(i + 1);
}
return ans;
}
}
```
Expand All @@ -261,46 +275,43 @@ class BinaryIndexedTree {

```cpp
class BinaryIndexedTree {
public:
private:
int n;
vector<int> c;

BinaryIndexedTree(int _n)
: n(_n)
, c(_n + 1) {}
public:
BinaryIndexedTree(int n)
: n(n)
, c(n + 1) {}

void update(int x, int delta) {
while (x <= n) {
for (; x <= n; x += x & -x) {
c[x] += delta;
x += lowbit(x);
}
}

int query(int x) {
int s = 0;
while (x > 0) {
for (; x > 0; x -= x & -x) {
s += c[x];
x -= lowbit(x);
}
return s;
}

int lowbit(int x) {
return x & -x;
}
};

class Solution {
public:
vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
BinaryIndexedTree* tree = new BinaryIndexedTree(length);
for (auto& e : updates) {
int start = e[0], end = e[1], inc = e[2];
tree->update(start + 1, inc);
tree->update(end + 2, -inc);
for (const auto& e : updates) {
int l = e[0], r = e[1], c = e[2];
tree->update(l + 1, c);
tree->update(r + 2, -c);
}
vector<int> ans;
for (int i = 0; i < length; ++i) ans.push_back(tree->query(i + 1));
for (int i = 0; i < length; ++i) {
ans.push_back(tree->query(i + 1));
}
return ans;
}
};
Expand All @@ -314,46 +325,116 @@ type BinaryIndexedTree struct {
c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
c := make([]int, n+1)
return &BinaryIndexedTree{n, c}
func NewBinaryIndexedTree(n int) *BinaryIndexedTree {
return &BinaryIndexedTree{n: n, c: make([]int, n+1)}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
for x <= this.n {
this.c[x] += delta
x += this.lowbit(x)
func (bit *BinaryIndexedTree) update(x, delta int) {
for ; x <= bit.n; x += x & -x {
bit.c[x] += delta
}
}

func (this *BinaryIndexedTree) query(x int) int {
func (bit *BinaryIndexedTree) query(x int) int {
s := 0
for x > 0 {
s += this.c[x]
x -= this.lowbit(x)
for ; x > 0; x -= x & -x {
s += bit.c[x]
}
return s
}

func getModifiedArray(length int, updates [][]int) []int {
tree := newBinaryIndexedTree(length)
func getModifiedArray(length int, updates [][]int) (ans []int) {
bit := NewBinaryIndexedTree(length)
for _, e := range updates {
start, end, inc := e[0], e[1], e[2]
tree.update(start+1, inc)
tree.update(end+2, -inc)
l, r, c := e[0], e[1], e[2]
bit.update(l+1, c)
bit.update(r+2, -c)
}
ans := make([]int, length)
for i := range ans {
ans[i] = tree.query(i + 1)
for i := 1; i <= length; i++ {
ans = append(ans, bit.query(i))
}
return ans
return
}
```

#### TypeScript

```ts
class BinaryIndexedTree {
private n: number;
private c: number[];

constructor(n: number) {
this.n = n;
this.c = Array(n + 1).fill(0);
}

update(x: number, delta: number): void {
for (; x <= this.n; x += x & -x) {
this.c[x] += delta;
}
}

query(x: number): number {
let s = 0;
for (; x > 0; x -= x & -x) {
s += this.c[x];
}
return s;
}
}

function getModifiedArray(length: number, updates: number[][]): number[] {
const bit = new BinaryIndexedTree(length);
for (const [l, r, c] of updates) {
bit.update(l + 1, c);
bit.update(r + 2, -c);
}
return Array.from({ length }, (_, i) => bit.query(i + 1));
}
```

#### JavaScript

```js
/**
* @param {number} length
* @param {number[][]} updates
* @return {number[]}
*/
var getModifiedArray = function (length, updates) {
class BinaryIndexedTree {
constructor(n) {
this.n = n;
this.c = Array(n + 1).fill(0);
}

update(x, delta) {
while (x <= this.n) {
this.c[x] += delta;
x += x & -x;
}
}

query(x) {
let s = 0;
while (x > 0) {
s += this.c[x];
x -= x & -x;
}
return s;
}
}

const bit = new BinaryIndexedTree(length);
for (const [l, r, c] of updates) {
bit.update(l + 1, c);
bit.update(r + 2, -c);
}
return Array.from({ length }, (_, i) => bit.query(i + 1));
};
```

<!-- tabs:end -->

<!-- solution:end -->
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