feat: add solutions to lc problem: No.2414 (#3486)

No.2414.Length of the Longest Alphabetical Continuous Substring

Author: yanglbme

Diff for: solution/2400-2499/2414.Length of the Longest Alphabetical Continuous Substring/README.md

@@ -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 $O(1)$。
 
@@ -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
 ```
 
@@ -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;
     }
 }
@@ -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;
     }
 };
@@ -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
 }
 ```
@@ -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;
 }
 ```
 
@@ -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
     }
 }
 ```
@@ -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;
 }
 ```
 

Diff for: solution/2400-2499/2414.Length of the Longest Alphabetical Continuous Substring/README_EN.md

@@ -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 $O(1)$.
 
@@ -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
 ```
 
@@ -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;
     }
 }
@@ -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;
     }
 };
@@ -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
 }
 ```
@@ -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;
 }
 ```
 
@@ -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
     }
 }
 ```
@@ -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;
 }
 ```
 

Diff for: solution/2400-2499/2414.Length of the Longest Alphabetical Continuous Substring/Solution.c

@@ -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;
+}