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Jun 4, 2024
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feat: update solutions to lc problems: No.412,416 #3022

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72 changes: 46 additions & 26 deletions solution/0400-0499/0412.Fizz Buzz/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -63,7 +63,11 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：模拟

我们遍历从 1 到 n 的每个整数，对于每个整数，我们检查它是否是 3 和 5 的倍数，或者只是 3 的倍数，或者只是 5 的倍数。根据检查的结果，我们将相应的字符串添加到答案数组中。

时间复杂度 $O(n)$，其中 $n$ 是题目给定的整数。忽略答案数组的空间消耗，空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand Down Expand Up @@ -118,9 +122,15 @@ public:
vector<string> ans;
for (int i = 1; i <= n; ++i) {
string s = "";
if (i % 3 == 0) s += "Fizz";
if (i % 5 == 0) s += "Buzz";
if (s.size() == 0) s = to_string(i);
if (i % 3 == 0) {
s += "Fizz";
}
if (i % 5 == 0) {
s += "Buzz";
}
if (s.empty()) {
s = to_string(i);
}
ans.push_back(s);
}
return ans;
Expand All @@ -131,8 +141,7 @@ public:
#### Go

```go
func fizzBuzz(n int) []string {
var ans []string
func fizzBuzz(n int) (ans []string) {
for i := 1; i <= n; i++ {
s := &strings.Builder{}
if i%3 == 0 {
Expand All @@ -146,22 +155,31 @@ func fizzBuzz(n int) []string {
}
ans = append(ans, s.String())
}
return ans
return
}
```

#### JavaScript

```js
const fizzBuzz = function (n) {
let arr = [];
for (let i = 1; i <= n; i++) {
if (i % 15 === 0) arr.push('FizzBuzz');
else if (i % 3 === 0) arr.push('Fizz');
else if (i % 5 === 0) arr.push('Buzz');
else arr.push(`${i}`);
/**
* @param {number} n
* @return {string[]}
*/
var fizzBuzz = function (n) {
const ans = [];
for (let i = 1; i <= n; ++i) {
if (i % 15 === 0) {
ans.push('FizzBuzz');
} else if (i % 3 === 0) {
ans.push('Fizz');
} else if (i % 5 === 0) {
ans.push('Buzz');
} else {
ans.push(`${i}`);
}
}
return arr;
return ans;
};
```

Expand All @@ -174,19 +192,21 @@ class Solution {
* @return String[]
*/
function fizzBuzz($n) {
$rs = [];
for ($i = 1; $i <= $n; $i++) {
if ($i % 3 != 0 && $i % 5 != 0) {
array_push($rs, strval($i));
} elseif ($i % 3 == 0 && $i % 5 != 0) {
array_push($rs, 'Fizz');
} elseif ($i % 3 != 0 && $i % 5 == 0) {
array_push($rs, 'Buzz');
} else {
array_push($rs, 'FizzBuzz');
$ans = [];
for ($i = 1; $i <= $n; ++$i) {
$s = '';
if ($i % 3 == 0) {
$s .= 'Fizz';
}
if ($i % 5 == 0) {
$s .= 'Buzz';
}
if (strlen($s) == 0) {
$s .= $i;
}
$ans[] = $s;
}
return $rs;
return $ans;
}
}
```
Expand Down
72 changes: 46 additions & 26 deletions solution/0400-0499/0412.Fizz Buzz/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -51,7 +51,11 @@ tags:

<!-- solution:start -->

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

We iterate through each integer from 1 to $n$. For each integer, we check whether it is a multiple of both 3 and 5, or just a multiple of 3, or just a multiple of 5. Based on the check result, we add the corresponding string to the answer array.

The time complexity is $O(n)$, where $n$ is the integer given in the problem. Ignoring the space consumption of the answer array, the space complexity is $O(1)$.

<!-- tabs:start -->

Expand Down Expand Up @@ -106,9 +110,15 @@ public:
vector<string> ans;
for (int i = 1; i <= n; ++i) {
string s = "";
if (i % 3 == 0) s += "Fizz";
if (i % 5 == 0) s += "Buzz";
if (s.size() == 0) s = to_string(i);
if (i % 3 == 0) {
s += "Fizz";
}
if (i % 5 == 0) {
s += "Buzz";
}
if (s.empty()) {
s = to_string(i);
}
ans.push_back(s);
}
return ans;
Expand All @@ -119,8 +129,7 @@ public:
#### Go

```go
func fizzBuzz(n int) []string {
var ans []string
func fizzBuzz(n int) (ans []string) {
for i := 1; i <= n; i++ {
s := &strings.Builder{}
if i%3 == 0 {
Expand All @@ -134,22 +143,31 @@ func fizzBuzz(n int) []string {
}
ans = append(ans, s.String())
}
return ans
return
}
```

#### JavaScript

```js
const fizzBuzz = function (n) {
let arr = [];
for (let i = 1; i <= n; i++) {
if (i % 15 === 0) arr.push('FizzBuzz');
else if (i % 3 === 0) arr.push('Fizz');
else if (i % 5 === 0) arr.push('Buzz');
else arr.push(`${i}`);
/**
* @param {number} n
* @return {string[]}
*/
var fizzBuzz = function (n) {
const ans = [];
for (let i = 1; i <= n; ++i) {
if (i % 15 === 0) {
ans.push('FizzBuzz');
} else if (i % 3 === 0) {
ans.push('Fizz');
} else if (i % 5 === 0) {
ans.push('Buzz');
} else {
ans.push(`${i}`);
}
}
return arr;
return ans;
};
```

Expand All @@ -162,19 +180,21 @@ class Solution {
* @return String[]
*/
function fizzBuzz($n) {
$rs = [];
for ($i = 1; $i <= $n; $i++) {
if ($i % 3 != 0 && $i % 5 != 0) {
array_push($rs, strval($i));
} elseif ($i % 3 == 0 && $i % 5 != 0) {
array_push($rs, 'Fizz');
} elseif ($i % 3 != 0 && $i % 5 == 0) {
array_push($rs, 'Buzz');
} else {
array_push($rs, 'FizzBuzz');
$ans = [];
for ($i = 1; $i <= $n; ++$i) {
$s = '';
if ($i % 3 == 0) {
$s .= 'Fizz';
}
if ($i % 5 == 0) {
$s .= 'Buzz';
}
if (strlen($s) == 0) {
$s .= $i;
}
$ans[] = $s;
}
return $rs;
return $ans;
}
}
```
Expand Down
12 changes: 9 additions & 3 deletions solution/0400-0499/0412.Fizz Buzz/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -4,9 +4,15 @@ class Solution {
vector<string> ans;
for (int i = 1; i <= n; ++i) {
string s = "";
if (i % 3 == 0) s += "Fizz";
if (i % 5 == 0) s += "Buzz";
if (s.size() == 0) s = to_string(i);
if (i % 3 == 0) {
s += "Fizz";
}
if (i % 5 == 0) {
s += "Buzz";
}
if (s.empty()) {
s = to_string(i);
}
ans.push_back(s);
}
return ans;
Expand Down
5 changes: 2 additions & 3 deletions solution/0400-0499/0412.Fizz Buzz/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,5 +1,4 @@
func fizzBuzz(n int) []string {
var ans []string
func fizzBuzz(n int) (ans []string) {
for i := 1; i <= n; i++ {
s := &strings.Builder{}
if i%3 == 0 {
Expand All @@ -13,5 +12,5 @@ func fizzBuzz(n int) []string {
}
ans = append(ans, s.String())
}
return ans
return
}
25 changes: 17 additions & 8 deletions solution/0400-0499/0412.Fizz Buzz/Solution.js
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Original file line number Diff line number Diff line change
@@ -1,10 +1,19 @@
const fizzBuzz = function (n) {
let arr = [];
for (let i = 1; i <= n; i++) {
if (i % 15 === 0) arr.push('FizzBuzz');
else if (i % 3 === 0) arr.push('Fizz');
else if (i % 5 === 0) arr.push('Buzz');
else arr.push(`${i}`);
/**
* @param {number} n
* @return {string[]}
*/
var fizzBuzz = function (n) {
const ans = [];
for (let i = 1; i <= n; ++i) {
if (i % 15 === 0) {
ans.push('FizzBuzz');
} else if (i % 3 === 0) {
ans.push('Fizz');
} else if (i % 5 === 0) {
ans.push('Buzz');
} else {
ans.push(`${i}`);
}
}
return arr;
return ans;
};
26 changes: 14 additions & 12 deletions solution/0400-0499/0412.Fizz Buzz/Solution.php
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Original file line number Diff line number Diff line change
Expand Up @@ -4,18 +4,20 @@ class Solution {
* @return String[]
*/
function fizzBuzz($n) {
$rs = [];
for ($i = 1; $i <= $n; $i++) {
if ($i % 3 != 0 && $i % 5 != 0) {
array_push($rs, strval($i));
} elseif ($i % 3 == 0 && $i % 5 != 0) {
array_push($rs, 'Fizz');
} elseif ($i % 3 != 0 && $i % 5 == 0) {
array_push($rs, 'Buzz');
} else {
array_push($rs, 'FizzBuzz');
$ans = [];
for ($i = 1; $i <= $n; ++$i) {
$s = '';
if ($i % 3 == 0) {
$s .= 'Fizz';
}
if ($i % 5 == 0) {
$s .= 'Buzz';
}
if (strlen($s) == 0) {
$s .= $i;
}
$ans[] = $s;
}
return $rs;
return $ans;
}
}
}
35 changes: 4 additions & 31 deletions solution/0400-0499/0413.Arithmetic Slices/README.md
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Expand Up @@ -67,7 +67,7 @@ tags:

我们用 $d$ 表示当前相邻两个元素的差值，用 $cnt$ 表示当前等差数列的长度，初始时 $d = 3000$, $cnt = 2$。

遍历数组 `nums`，对于相邻的两个元素 $a$ 和 $b$，如果 $b - a = d$，则说明当前元素 $b$ 也属于当前等差数列，此时 $cnt$ 自增 1；否则说明当前元素 $b$ 不属于当前等差数列，此时更新 $d = b - a$，$cnt = 2$。如果 $cnt \ge 3$，则说明当前等差数列的长度至少为 3，此时等差数列的个数为 $cnt - 2$，将其加到答案中。
遍历数组 `nums`，对于相邻的两个元素 $a$ 和 $b$，如果 $b - a = d$，则说明当前元素 $b$ 也属于当前等差数列，此时 $cnt$ 自增 1；否则说明当前元素 $b$ 不属于当前等差数列，此时更新 $d = b - a$，且 $cnt = 2$。如果 $cnt \ge 3$，则说明当前等差数列的长度至少为 3，此时等差数列的个数为 $cnt - 2$，将其加到答案中。

遍历结束后，即可得到答案。

Expand All @@ -87,15 +87,15 @@ tags:
```python
class Solution:
def numberOfArithmeticSlices(self, nums: List[int]) -> int:
ans, cnt = 0, 2
ans = cnt = 0
d = 3000
for a, b in pairwise(nums):
if b - a == d:
cnt += 1
else:
d = b - a
cnt = 2
ans += max(0, cnt - 2)
cnt = 0
ans += cnt
return ans
```

Expand Down Expand Up @@ -187,31 +187,4 @@ function numberOfArithmeticSlices(nums: number[]): number {

<!-- solution:end -->

<!-- solution:start -->

### 方法二

<!-- tabs:start -->

#### Python3

```python
class Solution:
def numberOfArithmeticSlices(self, nums: List[int]) -> int:
ans = cnt = 0
d = 3000
for a, b in pairwise(nums):
if b - a == d:
cnt += 1
else:
d = b - a
cnt = 0
ans += cnt
return ans
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->
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