feat: add solutions for lc No.2574

Author: yanglbme

solution/2500-2599/2573.Find the String with LCP/README.md

@@ -48,7 +48,7 @@ tags:
 <pre>
 <strong>输入：</strong>lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
 <strong>输出：</strong>"aaaa"
-<strong>解释：</strong>lcp 对应只有一个不同字母的任意 4 字母字符串，字典序最小的是 "aaaa" 。 
+<strong>解释：</strong>lcp 对应只有一个不同字母的任意 4 字母字符串，字典序最小的是 "aaaa" 。
 </pre>
 
 <p><strong>示例 3：</strong></p>
@@ -298,6 +298,56 @@ function findTheString(lcp: number[][]): string {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
+        let n = lcp.len();
+        let mut s = vec!['\0'; n];
+        let mut i = 0;
+
+        for c in b'a'..=b'z' {
+            while i < n && s[i] != '\0' {
+                i += 1;
+            }
+            if i == n {
+                break;
+            }
+            for j in i..n {
+                if lcp[i][j] > 0 {
+                    s[j] = c as char;
+                }
+            }
+        }
+
+        for i in 0..n {
+            if s[i] == '\0' {
+                return "".to_string();
+            }
+        }
+
+        for i in (0..n).rev() {
+            for j in (0..n).rev() {
+                if s[i] == s[j] {
+                    if i == n - 1 || j == n - 1 {
+                        if lcp[i][j] != 1 {
+                            return "".to_string();
+                        }
+                    } else if lcp[i][j] != lcp[i + 1][j + 1] + 1 {
+                        return "".to_string();
+                    }
+                } else if lcp[i][j] > 0 {
+                    return "".to_string();
+                }
+            }
+        }
+
+        s.into_iter().collect()
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2500-2599/2573.Find the String with LCP/README_EN.md

@@ -47,7 +47,7 @@ tags:
 <pre>
 <strong>Input:</strong> lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]
 <strong>Output:</strong> &quot;aaaa&quot;
-<strong>Explanation:</strong> lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is &quot;aaaa&quot;. 
+<strong>Explanation:</strong> lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is &quot;aaaa&quot;.
 </pre>
 
 <p><strong class="example">Example 3:</strong></p>
@@ -296,6 +296,56 @@ function findTheString(lcp: number[][]): string {
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
+        let n = lcp.len();
+        let mut s = vec!['\0'; n];
+        let mut i = 0;
+
+        for c in b'a'..=b'z' {
+            while i < n && s[i] != '\0' {
+                i += 1;
+            }
+            if i == n {
+                break;
+            }
+            for j in i..n {
+                if lcp[i][j] > 0 {
+                    s[j] = c as char;
+                }
+            }
+        }
+
+        for i in 0..n {
+            if s[i] == '\0' {
+                return "".to_string();
+            }
+        }
+
+        for i in (0..n).rev() {
+            for j in (0..n).rev() {
+                if s[i] == s[j] {
+                    if i == n - 1 || j == n - 1 {
+                        if lcp[i][j] != 1 {
+                            return "".to_string();
+                        }
+                    } else if lcp[i][j] != lcp[i + 1][j + 1] + 1 {
+                        return "".to_string();
+                    }
+                } else if lcp[i][j] > 0 {
+                    return "".to_string();
+                }
+            }
+        }
+
+        s.into_iter().collect()
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2500-2599/2573.Find the String with LCP/Solution.rs

@@ -0,0 +1,45 @@
+impl Solution {
+    pub fn find_the_string(lcp: Vec<Vec<i32>>) -> String {
+        let n = lcp.len();
+        let mut s = vec!['\0'; n];
+        let mut i = 0;
+
+        for c in b'a'..=b'z' {
+            while i < n && s[i] != '\0' {
+                i += 1;
+            }
+            if i == n {
+                break;
+            }
+            for j in i..n {
+                if lcp[i][j] > 0 {
+                    s[j] = c as char;
+                }
+            }
+        }
+
+        for i in 0..n {
+            if s[i] == '\0' {
+                return "".to_string();
+            }
+        }
+
+        for i in (0..n).rev() {
+            for j in (0..n).rev() {
+                if s[i] == s[j] {
+                    if i == n - 1 || j == n - 1 {
+                        if lcp[i][j] != 1 {
+                            return "".to_string();
+                        }
+                    } else if lcp[i][j] != lcp[i + 1][j + 1] + 1 {
+                        return "".to_string();
+                    }
+                } else if lcp[i][j] > 0 {
+                    return "".to_string();
+                }
+            }
+        }
+
+        s.into_iter().collect()
+    }
+}