diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md
index ec8ad61c..439ad8ea 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -587,5 +587,57 @@ func partition(_ s: String) -> [[String]] {
 }
 ```
 
+## Rust
+
+```rust
+impl Solution {
+    pub fn partition(s: String) -> Vec<Vec<String>> {
+        let mut ret = vec![];
+        let mut path = vec![];
+        let sub_str: Vec<char> = s.chars().collect();
+        
+        Self::backtracing(&sub_str, 0, &mut ret, &mut path);
+        
+        ret
+    }
+    
+    fn backtracing(sub_str: &Vec<char>, start: usize, ret: &mut Vec<Vec<String>>, path: &mut Vec<String>) {
+        //如果起始位置大于s的大小，说明找到了一组分割方案
+        if start >= sub_str.len() {
+            ret.push(path.clone());
+            return;
+        }
+        
+        for i in start..sub_str.len() {
+            if !Self::is_palindrome(sub_str, start, i) {
+                continue;
+            }
+            //如果是回文子串，则记录
+            let s: String = sub_str[start..i+1].into_iter().collect();
+            path.push(s);
+            
+            //起始位置后移，保证不重复
+            Self::backtracing(sub_str, i+1, ret, path); 
+            path.pop();
+        }
+        
+    }
+    
+    fn is_palindrome(s: &Vec<char>, start: usize, end: usize) -> bool {
+        let (mut start, mut end) = (start, end);
+        
+        while start < end {
+            if s[start] != s[end] {
+                return false;
+            }
+            
+            start += 1;
+            end -= 1;
+        }
+
+        true
+    }
+}
+```
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>