feat: add solutions to lc problem: No.0467

- No.0467.Unique Substrings in Wraparound String

Author: YangFong

solution/0400-0499/0467.Unique Substrings in Wraparound String/README.md

@@ -53,6 +53,10 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+题目导读，为什么示例 2 只有两个字串？
+
+> 成为字串的一个标准，需要是连续的，`a` 与 `c` 之间少了一个 `b`，所以不能算一个子字符串
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -71,6 +75,51 @@
 
 ```
 
+### **TypeScript**
+
+```ts
+function findSubstringInWraproundString(p: string): number {
+    const n = p.length;
+    const dp = new Array(26).fill(0);
+    let cur = 1;
+    dp[p.charCodeAt(0) - 'a'.charCodeAt(0)] = 1;
+    for (let i = 1; i < n; i++) {
+        if ((p.charCodeAt(i) - p.charCodeAt(i - 1) + 25) % 26 == 0) {
+            cur++;
+        } else {
+            cur = 1;
+        }
+        const index = p.charCodeAt(i) - 'a'.charCodeAt(0);
+        dp[index] = Math.max(dp[index], cur);
+    }
+    return dp.reduce((r, v) => r + v);
+}
+```
+
+### **Rust**
+
+```rust
+impl Solution {
+    pub fn find_substring_in_wrapround_string(p: String) -> i32 {
+        let n = p.len();
+        let p = p.as_bytes();
+        let mut dp = [0; 26];
+        let mut cur = 1;
+        dp[(p[0] - b'a') as usize] = 1;
+        for i in 1..n {
+            if (p[i] - p[i - 1] + 25) % 26 == 0 {
+                cur += 1;
+            } else {
+                cur = 1;
+            }
+            let index = (p[i] - b'a') as usize;
+            dp[index] = dp[index].max(cur);
+        }
+        dp.into_iter().sum()
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0400-0499/0467.Unique Substrings in Wraparound String/README_EN.md

@@ -61,6 +61,51 @@ Explanation: Only the substring "a" of p is in s.
 
 ```
 
+### **TypeScript**
+
+```ts
+function findSubstringInWraproundString(p: string): number {
+    const n = p.length;
+    const dp = new Array(26).fill(0);
+    let cur = 1;
+    dp[p.charCodeAt(0) - 'a'.charCodeAt(0)] = 1;
+    for (let i = 1; i < n; i++) {
+        if ((p.charCodeAt(i) - p.charCodeAt(i - 1) + 25) % 26 == 0) {
+            cur++;
+        } else {
+            cur = 1;
+        }
+        const index = p.charCodeAt(i) - 'a'.charCodeAt(0);
+        dp[index] = Math.max(dp[index], cur);
+    }
+    return dp.reduce((r, v) => r + v);
+}
+```
+
+### **Rust**
+
+```rust
+impl Solution {
+    pub fn find_substring_in_wrapround_string(p: String) -> i32 {
+        let n = p.len();
+        let p = p.as_bytes();
+        let mut dp = [0; 26];
+        let mut cur = 1;
+        dp[(p[0] - b'a') as usize] = 1;
+        for i in 1..n {
+            if (p[i] - p[i - 1] + 25) % 26 == 0 {
+                cur += 1;
+            } else {
+                cur = 1;
+            }
+            let index = (p[i] - b'a') as usize;
+            dp[index] = dp[index].max(cur);
+        }
+        dp.into_iter().sum()
+    }
+}
+```
+
 ### **...**
 
 ```

solution/0400-0499/0467.Unique Substrings in Wraparound String/Solution.rs

@@ -0,0 +1,19 @@
+impl Solution {
+    pub fn find_substring_in_wrapround_string(p: String) -> i32 {
+        let n = p.len();
+        let p = p.as_bytes();
+        let mut dp = [0; 26];
+        let mut cur = 1;
+        dp[(p[0] - b'a') as usize] = 1;
+        for i in 1..n {
+            if (p[i] - p[i - 1] + 25) % 26 == 0 {
+                cur += 1;
+            } else {
+                cur = 1;
+            }
+            let index = (p[i] - b'a') as usize;
+            dp[index] = dp[index].max(cur);
+        }
+        dp.into_iter().sum()
+    }
+}

solution/0400-0499/0467.Unique Substrings in Wraparound String/Solution.ts

@@ -0,0 +1,16 @@
+function findSubstringInWraproundString(p: string): number {
+    const n = p.length;
+    const dp = new Array(26).fill(0);
+    let cur = 1;
+    dp[p.charCodeAt(0) - 'a'.charCodeAt(0)] = 1;
+    for (let i = 1; i < n; i++) {
+        if ((p.charCodeAt(i) - p.charCodeAt(i - 1) + 25) % 26 == 0) {
+            cur++;
+        } else {
+            cur = 1;
+        }
+        const index = p.charCodeAt(i) - 'a'.charCodeAt(0);
+        dp[index] = Math.max(dp[index], cur);
+    }
+    return dp.reduce((r, v) => r + v);
+}