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feat: add solutions to lc problem: No.2414
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No.2414.Length of the Longest Alphabetical Continuous Substring
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@yanglbme
yanglbme committed Sep 3, 2024
1 parent 06b41f9 commit 33a6eeb
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110 changes: 56 additions & 54 deletions ...400-2499/2414.Length of the Longest Alphabetical Continuous Substring/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -58,9 +58,13 @@ tags:

<!-- solution:start -->

### 方法一：双指针
### 方法一：一次遍历

我们用双指针 $i$ 和 $j$ 分别指向当前连续子字符串的起始位置和结束位置。遍历字符串 $s$，如果当前字符 $s[j]$ 比 $s[j-1]$ 大，则 $j$ 向右移动一位，否则更新 $i$ 为 $j$，并更新最长连续子字符串的长度。
我们可以遍历字符串 $s$，用一个变量 $\textit{ans}$ 记录最长的字母序连续子字符串的长度，用另一个变量 $\textit{cnt}$ 记录当前连续子字符串的长度。初始时 $\textit{ans} = \textit{cnt} = 1$。

接下来，我们从下标为 $1$ 的字符开始遍历字符串 $s$，对于每个字符 $s[i]$，如果 $s[i] - s[i - 1] = 1$，则说明当前字符和前一个字符是连续的，此时 $\textit{cnt} = \textit{cnt} + 1$，并更新 $\textit{ans} = \max(\textit{ans}, \textit{cnt})$；否则，说明当前字符和前一个字符不连续，此时 $\textit{cnt} = 1$。

最终返回 $\textit{ans}$ 即可。

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

Expand All @@ -71,14 +75,13 @@ tags:
```python
class Solution:
def longestContinuousSubstring(self, s: str) -> int:
ans = 0
i, j = 0, 1
while j < len(s):
ans = max(ans, j - i)
if ord(s[j]) - ord(s[j - 1]) != 1:
i = j
j += 1
ans = max(ans, j - i)
ans = cnt = 1
for x, y in pairwise(map(ord, s)):
if y - x == 1:
cnt += 1
ans = max(ans, cnt)
else:
cnt = 1
return ans
```

Expand All @@ -87,15 +90,14 @@ class Solution:
```java
class Solution {
public int longestContinuousSubstring(String s) {
int ans = 0;
int i = 0, j = 1;
for (; j < s.length(); ++j) {
ans = Math.max(ans, j - i);
if (s.charAt(j) - s.charAt(j - 1) != 1) {
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < s.length(); ++i) {
if (s.charAt(i) - s.charAt(i - 1) == 1) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 1;
}
}
ans = Math.max(ans, j - i);
return ans;
}
}
Expand All @@ -107,15 +109,14 @@ class Solution {
class Solution {
public:
int longestContinuousSubstring(string s) {
int ans = 0;
int i = 0, j = 1;
for (; j < s.size(); ++j) {
ans = max(ans, j - i);
if (s[j] - s[j - 1] != 1) {
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < s.size(); ++i) {
if (s[i] - s[i - 1] == 1) {
ans = max(ans, ++cnt);
} else {
cnt = 1;
}
}
ans = max(ans, j - i);
return ans;
}
};
Expand All @@ -125,15 +126,15 @@ public:
```go
func longestContinuousSubstring(s string) int {
ans := 0
i, j := 0, 1
for ; j < len(s); j++ {
ans = max(ans, j-i)
if s[j]-s[j-1] != 1 {
i = j
ans, cnt := 1, 1
for i := range s[1:] {
if s[i+1]-s[i] == 1 {
cnt++
ans = max(ans, cnt)
} else {
cnt = 1
}
}
ans = max(ans, j-i)
return ans
}
```
Expand All @@ -142,16 +143,15 @@ func longestContinuousSubstring(s string) int {

```ts
function longestContinuousSubstring(s: string): number {
const n = s.length;
let res = 1;
let i = 0;
for (let j = 1; j < n; j++) {
if (s[j].charCodeAt(0) - s[j - 1].charCodeAt(0) !== 1) {
res = Math.max(res, j - i);
i = j;
let [ans, cnt] = [1, 1];
for (let i = 1; i < s.length; ++i) {
if (s.charCodeAt(i) - s.charCodeAt(i - 1) === 1) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 1;
}
}
return Math.max(res, n - i);
return ans;
}
```

Expand All @@ -160,17 +160,18 @@ function longestContinuousSubstring(s: string): number {
```rust
impl Solution {
pub fn longest_continuous_substring(s: String) -> i32 {
let mut ans = 1;
let mut cnt = 1;
let s = s.as_bytes();
let n = s.len();
let mut res = 1;
let mut i = 0;
for j in 1..n {
if s[j] - s[j - 1] != 1 {
res = res.max(j - i);
i = j;
for i in 1..s.len() {
if s[i] - s[i - 1] == 1 {
cnt += 1;
ans = ans.max(cnt);
} else {
cnt = 1;
}
}
res.max(n - i) as i32
ans
}
}
```
Expand All @@ -182,15 +183,16 @@ impl Solution {

int longestContinuousSubstring(char* s) {
int n = strlen(s);
int i = 0;
int res = 1;
for (int j = 1; j < n; j++) {
if (s[j] - s[j - 1] != 1) {
res = max(res, j - i);
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < n; ++i) {
if (s[i] - s[i - 1] == 1) {
++cnt;
ans = max(ans, cnt);
} else {
cnt = 1;
}
}
return max(res, n - i);
return ans;
}
```
Expand Down
110 changes: 56 additions & 54 deletions ...-2499/2414.Length of the Longest Alphabetical Continuous Substring/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -58,9 +58,13 @@ tags:

<!-- solution:start -->

### Solution 1: Two Pointers
### Solution 1: Single Pass

We use two pointers $i$ and $j$ to point to the start and end of the current consecutive substring respectively. Traverse the string $s$, if the current character $s[j]$ is greater than $s[j-1]$, then move $j$ one step to the right, otherwise update $i$ to $j$, and update the length of the longest consecutive substring.
We can traverse the string $s$ and use a variable $\textit{ans}$ to record the length of the longest lexicographically consecutive substring, and another variable $\textit{cnt}$ to record the length of the current consecutive substring. Initially, $\textit{ans} = \textit{cnt} = 1$.

Next, we start traversing the string $s$ from the character at index $1$. For each character $s[i]$, if $s[i] - s[i - 1] = 1$, it means the current character and the previous character are consecutive. In this case, $\textit{cnt} = \textit{cnt} + 1$, and we update $\textit{ans} = \max(\textit{ans}, \textit{cnt})$. Otherwise, it means the current character and the previous character are not consecutive, so $\textit{cnt} = 1$.

Finally, we return $\textit{ans}$.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

Expand All @@ -71,14 +75,13 @@ The time complexity is $O(n)$, where $n$ is the length of the string $s$. The sp
```python
class Solution:
def longestContinuousSubstring(self, s: str) -> int:
ans = 0
i, j = 0, 1
while j < len(s):
ans = max(ans, j - i)
if ord(s[j]) - ord(s[j - 1]) != 1:
i = j
j += 1
ans = max(ans, j - i)
ans = cnt = 1
for x, y in pairwise(map(ord, s)):
if y - x == 1:
cnt += 1
ans = max(ans, cnt)
else:
cnt = 1
return ans
```

Expand All @@ -87,15 +90,14 @@ class Solution:
```java
class Solution {
public int longestContinuousSubstring(String s) {
int ans = 0;
int i = 0, j = 1;
for (; j < s.length(); ++j) {
ans = Math.max(ans, j - i);
if (s.charAt(j) - s.charAt(j - 1) != 1) {
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < s.length(); ++i) {
if (s.charAt(i) - s.charAt(i - 1) == 1) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 1;
}
}
ans = Math.max(ans, j - i);
return ans;
}
}
Expand All @@ -107,15 +109,14 @@ class Solution {
class Solution {
public:
int longestContinuousSubstring(string s) {
int ans = 0;
int i = 0, j = 1;
for (; j < s.size(); ++j) {
ans = max(ans, j - i);
if (s[j] - s[j - 1] != 1) {
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < s.size(); ++i) {
if (s[i] - s[i - 1] == 1) {
ans = max(ans, ++cnt);
} else {
cnt = 1;
}
}
ans = max(ans, j - i);
return ans;
}
};
Expand All @@ -125,15 +126,15 @@ public:
```go
func longestContinuousSubstring(s string) int {
ans := 0
i, j := 0, 1
for ; j < len(s); j++ {
ans = max(ans, j-i)
if s[j]-s[j-1] != 1 {
i = j
ans, cnt := 1, 1
for i := range s[1:] {
if s[i+1]-s[i] == 1 {
cnt++
ans = max(ans, cnt)
} else {
cnt = 1
}
}
ans = max(ans, j-i)
return ans
}
```
Expand All @@ -142,16 +143,15 @@ func longestContinuousSubstring(s string) int {

```ts
function longestContinuousSubstring(s: string): number {
const n = s.length;
let res = 1;
let i = 0;
for (let j = 1; j < n; j++) {
if (s[j].charCodeAt(0) - s[j - 1].charCodeAt(0) !== 1) {
res = Math.max(res, j - i);
i = j;
let [ans, cnt] = [1, 1];
for (let i = 1; i < s.length; ++i) {
if (s.charCodeAt(i) - s.charCodeAt(i - 1) === 1) {
ans = Math.max(ans, ++cnt);
} else {
cnt = 1;
}
}
return Math.max(res, n - i);
return ans;
}
```

Expand All @@ -160,17 +160,18 @@ function longestContinuousSubstring(s: string): number {
```rust
impl Solution {
pub fn longest_continuous_substring(s: String) -> i32 {
let mut ans = 1;
let mut cnt = 1;
let s = s.as_bytes();
let n = s.len();
let mut res = 1;
let mut i = 0;
for j in 1..n {
if s[j] - s[j - 1] != 1 {
res = res.max(j - i);
i = j;
for i in 1..s.len() {
if s[i] - s[i - 1] == 1 {
cnt += 1;
ans = ans.max(cnt);
} else {
cnt = 1;
}
}
res.max(n - i) as i32
ans
}
}
```
Expand All @@ -182,15 +183,16 @@ impl Solution {

int longestContinuousSubstring(char* s) {
int n = strlen(s);
int i = 0;
int res = 1;
for (int j = 1; j < n; j++) {
if (s[j] - s[j - 1] != 1) {
res = max(res, j - i);
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < n; ++i) {
if (s[i] - s[i - 1] == 1) {
++cnt;
ans = max(ans, cnt);
} else {
cnt = 1;
}
}
return max(res, n - i);
return ans;
}
```
Expand Down
17 changes: 9 additions & 8 deletions solution/2400-2499/2414.Length of the Longest Alphabetical Continuous Substring/Solution.c
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Original file line number Diff line number Diff line change
Expand Up @@ -2,13 +2,14 @@

int longestContinuousSubstring(char* s) {
int n = strlen(s);
int i = 0;
int res = 1;
for (int j = 1; j < n; j++) {
if (s[j] - s[j - 1] != 1) {
res = max(res, j - i);
i = j;
int ans = 1, cnt = 1;
for (int i = 1; i < n; ++i) {
if (s[i] - s[i - 1] == 1) {
++cnt;
ans = max(ans, cnt);
} else {
cnt = 1;
}
}
return max(res, n - i);
}
return ans;
}
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