Merge pull request #1760 from codeSu97/master

feat: 0070.爬楼梯.md 增加 rust代码

problems/0070.爬楼梯.md

@@ -28,7 +28,7 @@
     * 1 阶 + 2 阶
     * 2 阶 + 1 阶
 
-# 视频讲解 
+# 视频讲解
 
 **《代码随想录》算法视频公开课：[带你学透动态规划-爬楼梯|LeetCode:70.爬楼梯）](https://www.bilibili.com/video/BV17h411h7UH)，相信结合视频在看本篇题解，更有助于大家对本题的理解**。
 
@@ -216,7 +216,7 @@ public:
 ## 其他语言版本
 
 
-### Java 
+### Java
 
 ```java
 // 常规方式
@@ -237,7 +237,7 @@ class Solution {
     public int climbStairs(int n) {
         if(n <= 2) return n;
         int a = 1, b = 2, sum = 0;
-        
+
         for(int i = 3; i <= n; i++){
             sum = a + b;  // f(i - 1) + f(i - 2)
             a = b;        // 记录f(i - 1)，即下一轮的f(i - 2)
@@ -261,7 +261,7 @@ class Solution:
         for i in range(2,n+1):
             dp[i]=dp[i-1]+dp[i-2]
         return dp[n]
-        
+
 # 空间复杂度为O(1)版本
 class Solution:
     def climbStairs(self, n: int) -> int:
@@ -275,7 +275,7 @@ class Solution:
         return dp[1]
 ```
 
-### Go 
+### Go
 ```Go
 func climbStairs(n int) int {
     if n==1{
@@ -303,7 +303,7 @@ var climbStairs = function(n) {
 };
 ```
 
-TypeScript
+### TypeScript
 
 > 爬2阶
 
@@ -447,7 +447,26 @@ public class Solution {
 }
 ```
 
+### Rust
 
+```rust
+impl Solution {
+    pub fn climb_stairs(n: i32) -> i32 {
+        if n <= 2 {
+            return n;
+        }
+        let mut a = 1;
+        let mut b = 2;
+        let mut f = 0;
+        for i in 2..n {
+            f = a + b;
+            a = b;
+            b = f;
+        }
+        return f;
+    }
+}
+```
 
 <p align="center">
 <a href="https://programmercarl.com/other/kstar.html" target="_blank">