添加 0131.分割回文串 Rust回溯+动态规划版本

添加 0131.分割回文串 Rust回溯+动态规划版本

problems/0131.分割回文串.md

@@ -626,7 +626,8 @@ func partition(_ s: String) -> [[String]] {
 
 ## Rust
 
-```rust
+**回溯+函数判断回文串**
+```Rust
 impl Solution {
     pub fn partition(s: String) -> Vec<Vec<String>> {
         let mut ret = vec![];
@@ -676,6 +677,40 @@ impl Solution {
     }
 }
 ```
+**回溯+动态规划预处理判断回文串**
+```Rust
+impl Solution {
+    pub fn backtracking(is_palindrome: &Vec<Vec<bool>>, result: &mut Vec<Vec<String>>, path: &mut Vec<String>, s: &Vec<char>, start_index: usize) {
+        let len = s.len();
+        if start_index >= len {
+            result.push(path.to_vec());
+            return;
+        }
+        for i in start_index..len {
+            if is_palindrome[start_index][i] { path.push(s[start_index..=i].iter().collect::<String>()); } else { continue; }
+            Self::backtracking(is_palindrome, result, path, s, i + 1);
+            path.pop();
+        }
+    }
+
+    pub fn partition(s: String) -> Vec<Vec<String>> {
+        let mut result: Vec<Vec<String>> = Vec::new();
+        let mut path: Vec<String> = Vec::new();
+        let s = s.chars().collect::<Vec<char>>();
+        let len: usize = s.len();
+	// 使用动态规划预先打表
+	// 当且仅当其为空串(i>j)，或其长度为1(i=j)，或者首尾字符相同且(s[i+1..j−1])时为回文串
+        let mut is_palindrome = vec![vec![true; len]; len];
+        for i in (0..len).rev() {
+            for j in (i + 1)..len {
+                is_palindrome[i][j] = s[i] == s[j] && is_palindrome[i + 1][j - 1];
+            }
+        }
+        Self::backtracking(&is_palindrome, &mut result, &mut path, &s, 0);
+        result
+    }
+}
+```
 
 
 ## Scala