题解 | #删除链表的节点# | Rust

/**
 *  #[derive(PartialEq, Eq, Debug, Clone)]
 *  pub struct ListNode {
 *      pub val: i32,
 *      pub next: Option<Box<ListNode>>
 *  }
 * 
 *  impl ListNode {
 *      #[inline]
 *      fn new(val: i32) -> Self {
 *          ListNode {
 *              val: val,
 *              next: None,
 *          }
 *      }
 *  }
 */
struct Solution{

}

impl Solution {
    fn new() -> Self {
        Solution{}
    }

    /**
    * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
    *
    * 
        * @param head ListNode类 
        * @param val int整型 
        * @return ListNode类
    */
    pub fn deleteNode(&self, head: Option<Box<ListNode>>, val: i32) -> Option<Box<ListNode>> {
        let mut ans = Box::new(ListNode::new(-1));
        ans.next = head;
        let mut cur = &mut ans;
        while cur.next.is_some() {
             if cur.next.as_ref().unwrap().val == val {
                cur.next = cur.next.as_mut().unwrap().next.take();
             } else {
                cur = cur.next.as_mut().unwrap();
             }
        }
        return ans.next;
    }
}