Merge two sorted lists in Rust











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I wrote the following code to merge two sorted lists. Is there a way I can improve it?



Possible ideas (not sure how to implement them):




  • Remove code duplication (setting the returning list node and moving to the next)

  • Avoid the use of the infinite loop

  • Avoid the use of panic!


This is the data structure:



type Link = Option<Box<Node>>;

pub struct Node {
elem: i32,
next: Link,
}

impl Node {
fn new(elem: i32) -> Node {
Node { elem, next: None }
}
}


And this is the method:



fn merge_sorted_lists(list1: &Link, list2: &Link) -> Link {
if list1.is_none() || list2.is_none() {
return None;
}
let mut res = None;
{
let mut node3 = &mut res;
let mut node1 = list1;
let mut node2 = list2;
loop {
if let (Some(link1), Some(link2)) = (node1, node2) {
if link1.elem > link2.elem {
*node3 = Some(Box::new(Node::new(link2.elem)));
node2 = &link2.next;
} else {
*node3 = Some(Box::new(Node::new(link1.elem)));
node1 = &link1.next;
}
if let Some(link) = {node3} {
node3 = &mut link.next;
} else {
panic!();
}
} else if let Some(link1) = node1 {
*node3 = Some(Box::new(Node::new(link1.elem)));
node1 = &link1.next;
if let Some(link) = {node3} {
node3 = &mut link.next;
} else {
panic!();
}
} else if let Some(link2) = node2 {
*node3 = Some(Box::new(Node::new(link2.elem)));
node1 = &link2.next;
if let Some(link) = {node3} {
node3 = &mut link.next;
} else {
panic!();
}
} else {
break;
}
}
}
res
}









share|improve this question
















bumped to the homepage by Community 17 hours ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.



















    up vote
    3
    down vote

    favorite












    I wrote the following code to merge two sorted lists. Is there a way I can improve it?



    Possible ideas (not sure how to implement them):




    • Remove code duplication (setting the returning list node and moving to the next)

    • Avoid the use of the infinite loop

    • Avoid the use of panic!


    This is the data structure:



    type Link = Option<Box<Node>>;

    pub struct Node {
    elem: i32,
    next: Link,
    }

    impl Node {
    fn new(elem: i32) -> Node {
    Node { elem, next: None }
    }
    }


    And this is the method:



    fn merge_sorted_lists(list1: &Link, list2: &Link) -> Link {
    if list1.is_none() || list2.is_none() {
    return None;
    }
    let mut res = None;
    {
    let mut node3 = &mut res;
    let mut node1 = list1;
    let mut node2 = list2;
    loop {
    if let (Some(link1), Some(link2)) = (node1, node2) {
    if link1.elem > link2.elem {
    *node3 = Some(Box::new(Node::new(link2.elem)));
    node2 = &link2.next;
    } else {
    *node3 = Some(Box::new(Node::new(link1.elem)));
    node1 = &link1.next;
    }
    if let Some(link) = {node3} {
    node3 = &mut link.next;
    } else {
    panic!();
    }
    } else if let Some(link1) = node1 {
    *node3 = Some(Box::new(Node::new(link1.elem)));
    node1 = &link1.next;
    if let Some(link) = {node3} {
    node3 = &mut link.next;
    } else {
    panic!();
    }
    } else if let Some(link2) = node2 {
    *node3 = Some(Box::new(Node::new(link2.elem)));
    node1 = &link2.next;
    if let Some(link) = {node3} {
    node3 = &mut link.next;
    } else {
    panic!();
    }
    } else {
    break;
    }
    }
    }
    res
    }









    share|improve this question
















    bumped to the homepage by Community 17 hours ago


    This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.

















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I wrote the following code to merge two sorted lists. Is there a way I can improve it?



      Possible ideas (not sure how to implement them):




      • Remove code duplication (setting the returning list node and moving to the next)

      • Avoid the use of the infinite loop

      • Avoid the use of panic!


      This is the data structure:



      type Link = Option<Box<Node>>;

      pub struct Node {
      elem: i32,
      next: Link,
      }

      impl Node {
      fn new(elem: i32) -> Node {
      Node { elem, next: None }
      }
      }


      And this is the method:



      fn merge_sorted_lists(list1: &Link, list2: &Link) -> Link {
      if list1.is_none() || list2.is_none() {
      return None;
      }
      let mut res = None;
      {
      let mut node3 = &mut res;
      let mut node1 = list1;
      let mut node2 = list2;
      loop {
      if let (Some(link1), Some(link2)) = (node1, node2) {
      if link1.elem > link2.elem {
      *node3 = Some(Box::new(Node::new(link2.elem)));
      node2 = &link2.next;
      } else {
      *node3 = Some(Box::new(Node::new(link1.elem)));
      node1 = &link1.next;
      }
      if let Some(link) = {node3} {
      node3 = &mut link.next;
      } else {
      panic!();
      }
      } else if let Some(link1) = node1 {
      *node3 = Some(Box::new(Node::new(link1.elem)));
      node1 = &link1.next;
      if let Some(link) = {node3} {
      node3 = &mut link.next;
      } else {
      panic!();
      }
      } else if let Some(link2) = node2 {
      *node3 = Some(Box::new(Node::new(link2.elem)));
      node1 = &link2.next;
      if let Some(link) = {node3} {
      node3 = &mut link.next;
      } else {
      panic!();
      }
      } else {
      break;
      }
      }
      }
      res
      }









      share|improve this question















      I wrote the following code to merge two sorted lists. Is there a way I can improve it?



      Possible ideas (not sure how to implement them):




      • Remove code duplication (setting the returning list node and moving to the next)

      • Avoid the use of the infinite loop

      • Avoid the use of panic!


      This is the data structure:



      type Link = Option<Box<Node>>;

      pub struct Node {
      elem: i32,
      next: Link,
      }

      impl Node {
      fn new(elem: i32) -> Node {
      Node { elem, next: None }
      }
      }


      And this is the method:



      fn merge_sorted_lists(list1: &Link, list2: &Link) -> Link {
      if list1.is_none() || list2.is_none() {
      return None;
      }
      let mut res = None;
      {
      let mut node3 = &mut res;
      let mut node1 = list1;
      let mut node2 = list2;
      loop {
      if let (Some(link1), Some(link2)) = (node1, node2) {
      if link1.elem > link2.elem {
      *node3 = Some(Box::new(Node::new(link2.elem)));
      node2 = &link2.next;
      } else {
      *node3 = Some(Box::new(Node::new(link1.elem)));
      node1 = &link1.next;
      }
      if let Some(link) = {node3} {
      node3 = &mut link.next;
      } else {
      panic!();
      }
      } else if let Some(link1) = node1 {
      *node3 = Some(Box::new(Node::new(link1.elem)));
      node1 = &link1.next;
      if let Some(link) = {node3} {
      node3 = &mut link.next;
      } else {
      panic!();
      }
      } else if let Some(link2) = node2 {
      *node3 = Some(Box::new(Node::new(link2.elem)));
      node1 = &link2.next;
      if let Some(link) = {node3} {
      node3 = &mut link.next;
      } else {
      panic!();
      }
      } else {
      break;
      }
      }
      }
      res
      }






      linked-list rust






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      edited Nov 11 at 12:11

























      asked Nov 11 at 11:41









      Nick

      14015




      14015





      bumped to the homepage by Community 17 hours ago


      This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







      bumped to the homepage by Community 17 hours ago


      This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
























          1 Answer
          1






          active

          oldest

          votes

















          up vote
          0
          down vote













          One problem in your current code is that you allocate the merged list, that could be what the user want but that not very flexible, a better way would be to consume the list, and let the user do a copy before if needed.



          One other problem is that you have spaghetti code, it's very difficult to maintain and error prone.



          you should also use generic to allow an user to have any type in your linked list.



          To solve these problem you could use recursion, the typical use case match yours, so I would simply rework all your code to use it:



          type Link<T> = Option<Box<Node<T>>>;

          #[derive(Debug, Clone)]
          pub struct Node<T> {
          elem: T,
          next: Link<T>,
          }

          impl<T> Node<T> {
          // rework of new to make it more flexible
          fn new(elem: T, next: Link<T>) -> Self {
          Self { elem, next }
          }

          // next allow to change to linked node and to return the old one
          fn next(&mut self, next: Link<T>) -> Link<T> {
          std::mem::replace(&mut self.next, next)
          }

          fn elem(&self) -> &T {
          &self.elem
          }

          // now we take by value and allow user to have a flexible control with f
          fn merge_by<F>(a: Link<T>, b: Link<T>, accu: Link<T>, f: F) -> Link<T>
          where
          F: Fn(&T, &T) -> bool,
          {
          match (a, b) {
          (Some(mut a), Some(mut b)) => {
          if f(a.elem(), b.elem()) {
          Self::merge_by(a.next(accu), Some(b), Some(a), f)
          } else {
          Self::merge_by(Some(a), b.next(accu), Some(b), f)
          }
          }
          (Some(a), None) => Self::rev(accu, Some(a)),
          (None, Some(b)) => Self::rev(accu, Some(b)),
          (None, None) => Self::rev(accu, None),
          }
          }

          // rev is needed when you deal with list
          fn rev(list: Link<T>, accu: Link<T>) -> Link<T> {
          match list {
          Some(mut list) => Self::rev(list.next(accu), Some(list)),
          None => accu,
          }
          }

          }

          fn main() {
          let a = Some(Box::new(Node::new(21, Some(Box::new(Node::new(42, None))))));
          println!("{:#?}", a);

          let b = Some(Box::new(Node::new(1, Some(Box::new(Node::new(2, None))))));
          println!("{:#?}", b);

          let c = Node::merge_by(a, b, None, std::cmp::PartialEq::eq);
          println!("{:#?}", c);
          }





          share|improve this answer























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            1 Answer
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            1 Answer
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            up vote
            0
            down vote













            One problem in your current code is that you allocate the merged list, that could be what the user want but that not very flexible, a better way would be to consume the list, and let the user do a copy before if needed.



            One other problem is that you have spaghetti code, it's very difficult to maintain and error prone.



            you should also use generic to allow an user to have any type in your linked list.



            To solve these problem you could use recursion, the typical use case match yours, so I would simply rework all your code to use it:



            type Link<T> = Option<Box<Node<T>>>;

            #[derive(Debug, Clone)]
            pub struct Node<T> {
            elem: T,
            next: Link<T>,
            }

            impl<T> Node<T> {
            // rework of new to make it more flexible
            fn new(elem: T, next: Link<T>) -> Self {
            Self { elem, next }
            }

            // next allow to change to linked node and to return the old one
            fn next(&mut self, next: Link<T>) -> Link<T> {
            std::mem::replace(&mut self.next, next)
            }

            fn elem(&self) -> &T {
            &self.elem
            }

            // now we take by value and allow user to have a flexible control with f
            fn merge_by<F>(a: Link<T>, b: Link<T>, accu: Link<T>, f: F) -> Link<T>
            where
            F: Fn(&T, &T) -> bool,
            {
            match (a, b) {
            (Some(mut a), Some(mut b)) => {
            if f(a.elem(), b.elem()) {
            Self::merge_by(a.next(accu), Some(b), Some(a), f)
            } else {
            Self::merge_by(Some(a), b.next(accu), Some(b), f)
            }
            }
            (Some(a), None) => Self::rev(accu, Some(a)),
            (None, Some(b)) => Self::rev(accu, Some(b)),
            (None, None) => Self::rev(accu, None),
            }
            }

            // rev is needed when you deal with list
            fn rev(list: Link<T>, accu: Link<T>) -> Link<T> {
            match list {
            Some(mut list) => Self::rev(list.next(accu), Some(list)),
            None => accu,
            }
            }

            }

            fn main() {
            let a = Some(Box::new(Node::new(21, Some(Box::new(Node::new(42, None))))));
            println!("{:#?}", a);

            let b = Some(Box::new(Node::new(1, Some(Box::new(Node::new(2, None))))));
            println!("{:#?}", b);

            let c = Node::merge_by(a, b, None, std::cmp::PartialEq::eq);
            println!("{:#?}", c);
            }





            share|improve this answer



























              up vote
              0
              down vote













              One problem in your current code is that you allocate the merged list, that could be what the user want but that not very flexible, a better way would be to consume the list, and let the user do a copy before if needed.



              One other problem is that you have spaghetti code, it's very difficult to maintain and error prone.



              you should also use generic to allow an user to have any type in your linked list.



              To solve these problem you could use recursion, the typical use case match yours, so I would simply rework all your code to use it:



              type Link<T> = Option<Box<Node<T>>>;

              #[derive(Debug, Clone)]
              pub struct Node<T> {
              elem: T,
              next: Link<T>,
              }

              impl<T> Node<T> {
              // rework of new to make it more flexible
              fn new(elem: T, next: Link<T>) -> Self {
              Self { elem, next }
              }

              // next allow to change to linked node and to return the old one
              fn next(&mut self, next: Link<T>) -> Link<T> {
              std::mem::replace(&mut self.next, next)
              }

              fn elem(&self) -> &T {
              &self.elem
              }

              // now we take by value and allow user to have a flexible control with f
              fn merge_by<F>(a: Link<T>, b: Link<T>, accu: Link<T>, f: F) -> Link<T>
              where
              F: Fn(&T, &T) -> bool,
              {
              match (a, b) {
              (Some(mut a), Some(mut b)) => {
              if f(a.elem(), b.elem()) {
              Self::merge_by(a.next(accu), Some(b), Some(a), f)
              } else {
              Self::merge_by(Some(a), b.next(accu), Some(b), f)
              }
              }
              (Some(a), None) => Self::rev(accu, Some(a)),
              (None, Some(b)) => Self::rev(accu, Some(b)),
              (None, None) => Self::rev(accu, None),
              }
              }

              // rev is needed when you deal with list
              fn rev(list: Link<T>, accu: Link<T>) -> Link<T> {
              match list {
              Some(mut list) => Self::rev(list.next(accu), Some(list)),
              None => accu,
              }
              }

              }

              fn main() {
              let a = Some(Box::new(Node::new(21, Some(Box::new(Node::new(42, None))))));
              println!("{:#?}", a);

              let b = Some(Box::new(Node::new(1, Some(Box::new(Node::new(2, None))))));
              println!("{:#?}", b);

              let c = Node::merge_by(a, b, None, std::cmp::PartialEq::eq);
              println!("{:#?}", c);
              }





              share|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                One problem in your current code is that you allocate the merged list, that could be what the user want but that not very flexible, a better way would be to consume the list, and let the user do a copy before if needed.



                One other problem is that you have spaghetti code, it's very difficult to maintain and error prone.



                you should also use generic to allow an user to have any type in your linked list.



                To solve these problem you could use recursion, the typical use case match yours, so I would simply rework all your code to use it:



                type Link<T> = Option<Box<Node<T>>>;

                #[derive(Debug, Clone)]
                pub struct Node<T> {
                elem: T,
                next: Link<T>,
                }

                impl<T> Node<T> {
                // rework of new to make it more flexible
                fn new(elem: T, next: Link<T>) -> Self {
                Self { elem, next }
                }

                // next allow to change to linked node and to return the old one
                fn next(&mut self, next: Link<T>) -> Link<T> {
                std::mem::replace(&mut self.next, next)
                }

                fn elem(&self) -> &T {
                &self.elem
                }

                // now we take by value and allow user to have a flexible control with f
                fn merge_by<F>(a: Link<T>, b: Link<T>, accu: Link<T>, f: F) -> Link<T>
                where
                F: Fn(&T, &T) -> bool,
                {
                match (a, b) {
                (Some(mut a), Some(mut b)) => {
                if f(a.elem(), b.elem()) {
                Self::merge_by(a.next(accu), Some(b), Some(a), f)
                } else {
                Self::merge_by(Some(a), b.next(accu), Some(b), f)
                }
                }
                (Some(a), None) => Self::rev(accu, Some(a)),
                (None, Some(b)) => Self::rev(accu, Some(b)),
                (None, None) => Self::rev(accu, None),
                }
                }

                // rev is needed when you deal with list
                fn rev(list: Link<T>, accu: Link<T>) -> Link<T> {
                match list {
                Some(mut list) => Self::rev(list.next(accu), Some(list)),
                None => accu,
                }
                }

                }

                fn main() {
                let a = Some(Box::new(Node::new(21, Some(Box::new(Node::new(42, None))))));
                println!("{:#?}", a);

                let b = Some(Box::new(Node::new(1, Some(Box::new(Node::new(2, None))))));
                println!("{:#?}", b);

                let c = Node::merge_by(a, b, None, std::cmp::PartialEq::eq);
                println!("{:#?}", c);
                }





                share|improve this answer














                One problem in your current code is that you allocate the merged list, that could be what the user want but that not very flexible, a better way would be to consume the list, and let the user do a copy before if needed.



                One other problem is that you have spaghetti code, it's very difficult to maintain and error prone.



                you should also use generic to allow an user to have any type in your linked list.



                To solve these problem you could use recursion, the typical use case match yours, so I would simply rework all your code to use it:



                type Link<T> = Option<Box<Node<T>>>;

                #[derive(Debug, Clone)]
                pub struct Node<T> {
                elem: T,
                next: Link<T>,
                }

                impl<T> Node<T> {
                // rework of new to make it more flexible
                fn new(elem: T, next: Link<T>) -> Self {
                Self { elem, next }
                }

                // next allow to change to linked node and to return the old one
                fn next(&mut self, next: Link<T>) -> Link<T> {
                std::mem::replace(&mut self.next, next)
                }

                fn elem(&self) -> &T {
                &self.elem
                }

                // now we take by value and allow user to have a flexible control with f
                fn merge_by<F>(a: Link<T>, b: Link<T>, accu: Link<T>, f: F) -> Link<T>
                where
                F: Fn(&T, &T) -> bool,
                {
                match (a, b) {
                (Some(mut a), Some(mut b)) => {
                if f(a.elem(), b.elem()) {
                Self::merge_by(a.next(accu), Some(b), Some(a), f)
                } else {
                Self::merge_by(Some(a), b.next(accu), Some(b), f)
                }
                }
                (Some(a), None) => Self::rev(accu, Some(a)),
                (None, Some(b)) => Self::rev(accu, Some(b)),
                (None, None) => Self::rev(accu, None),
                }
                }

                // rev is needed when you deal with list
                fn rev(list: Link<T>, accu: Link<T>) -> Link<T> {
                match list {
                Some(mut list) => Self::rev(list.next(accu), Some(list)),
                None => accu,
                }
                }

                }

                fn main() {
                let a = Some(Box::new(Node::new(21, Some(Box::new(Node::new(42, None))))));
                println!("{:#?}", a);

                let b = Some(Box::new(Node::new(1, Some(Box::new(Node::new(2, None))))));
                println!("{:#?}", b);

                let c = Node::merge_by(a, b, None, std::cmp::PartialEq::eq);
                println!("{:#?}", c);
                }






                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 11 at 15:22

























                answered Nov 11 at 14:21









                Stargateur

                21815




                21815






























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