题解 | #合并两个排序的链表# | 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 pHead1 ListNode类 
        * @param pHead2 ListNode类 
        * @return ListNode类
    */
    pub fn Merge(&self, pHead1: Option<Box<ListNode>>, pHead2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        // write code here
        if pHead1 == None {
            return pHead2;
        }
        if pHead2 == None {
            return pHead1;
        }
        let mut ans : Option<Box<ListNode>> = None;
        if pHead1.as_ref().unwrap().val < pHead2.as_ref().unwrap().val {
            ans = pHead1;
            ans.as_mut().unwrap().next = Solution::Merge(self, ans.as_mut().unwrap().next.take(), pHead2);
        } else {
            ans = pHead2;
            ans.as_mut().unwrap().next = Solution::Merge(self, pHead1, ans.as_mut().unwrap().next.take());
        }
        return ans;
    }
}