feat: add rust solution to lcof problem: No.22

面试题22. 链表中倒数第k个节点

Author: YangFong

lcof/面试题22. 链表中倒数第k个节点/README.md

@@ -137,6 +137,41 @@ public:
 };
 ```
 
+### **Rust**
+
+```rust
+// Definition for singly-linked list.
+// #[derive(PartialEq, Eq, Clone, Debug)]
+// pub struct ListNode {
+//   pub val: i32,
+//   pub next: Option<Box<ListNode>>
+// }
+// 
+// impl ListNode {
+//   #[inline]
+//   fn new(val: i32) -> Self {
+//     ListNode {
+//       next: None,
+//       val
+//     }
+//   }
+// }
+impl Solution {
+    pub fn get_kth_from_end(head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
+        let mut fast = &head; 
+        for _ in 0..k {
+            fast = &fast.as_ref().unwrap().next;
+        }
+        let mut slow = &head;
+        while let (Some(nf), Some(ns)) = (fast, slow) {
+            fast = &nf.next;
+            slow = &ns.next;
+        }
+        slow.to_owned()
+    }
+}
+```
+
 ### **...**
 
 ```

lcof/面试题22. 链表中倒数第k个节点/Solution.rs

@@ -0,0 +1,30 @@
+// Definition for singly-linked list.
+// #[derive(PartialEq, Eq, Clone, Debug)]
+// pub struct ListNode {
+//   pub val: i32,
+//   pub next: Option<Box<ListNode>>
+// }
+//
+// impl ListNode {
+//   #[inline]
+//   fn new(val: i32) -> Self {
+//     ListNode {
+//       next: None,
+//       val
+//     }
+//   }
+// }
+impl Solution {
+    pub fn get_kth_from_end(head: Option<Box<ListNode>>, k: i32) -> Option<Box<ListNode>> {
+        let mut fast = &head;
+        for _ in 0..k {
+            fast = &fast.as_ref().unwrap().next;
+        }
+        let mut slow = &head;
+        while let (Some(nf), Some(ns)) = (fast, slow) {
+            fast = &nf.next;
+            slow = &ns.next;
+        }
+        slow.to_owned()
+    }
+}