Question

面试题 25. 合并两个排序的链表

题目描述

输入两个递增排序的链表，合并这两个链表并使新链表中的节点仍然是递增排序的。

示例1：

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

限制：

0 <= 链表长度 <= 1000

注意：本题与主站 21 题相同：https://leetcode.cn/problems/merge-two-sorted-lists/

解法

方法一：迭代

我们先创建一个虚拟头结点 dummy，然后创建一个指针 cur 指向 dummy。

接下来，循环比较 l1 和 l2 的值，将较小的值接在 cur 后面，然后将指针向后移动一位。循环结束，将 cur 指向 l1 或者 l2 中剩余的部分。

最后返回 dummy.next 即可。

时间复杂度 $O(m + n)$，空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别为两个链表的长度。

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
    dummy = cur = ListNode(0)
    while l1 and l2:
      if l1.val <= l2.val:
        cur.next = l1
        l1 = l1.next
      else:
        cur.next = l2
        l2 = l2.next
      cur = cur.next
    cur.next = l1 or l2
    return dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode cur = dummy;
    while (l1 != null && l2 != null) {
      if (l1.val <= l2.val) {
        cur.next = l1;
        l1 = l1.next;
      } else {
        cur.next = l2;
        l2 = l2.next;
      }
      cur = cur.next;
    }
    cur.next = l1 == null ? l2 : l1;
    return dummy.next;
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    ListNode* dummy = new ListNode(0);
    ListNode* cur = dummy;
    while (l1 && l2) {
      if (l1->val <= l2->val) {
        cur->next = l1;
        l1 = l1->next;
      } else {
        cur->next = l2;
        l2 = l2->next;
      }
      cur = cur->next;
    }
    cur->next = l1 ? l1 : l2;
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
  dummy := &ListNode{}
  cur := dummy
  for l1 != nil && l2 != nil {
    if l1.Val <= l2.Val {
      cur.Next = l1
      l1 = l1.Next
    } else {
      cur.Next = l2
      l2 = l2.Next
    }
    cur = cur.Next
  }
  if l1 == nil {
    cur.Next = l2
  } else {
    cur.Next = l1
  }
  return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function mergeTwoLists(l1: ListNode | null, l2: ListNode | null): ListNode | null {
  const dummy = new ListNode(0);
  let cur = dummy;
  while (l1 && l2) {
    if (l1.val <= l2.val) {
      cur.next = l1;
      l1 = l1.next;
    } else {
      cur.next = l2;
      l2 = l2.next;
    }
    cur = cur.next;
  }
  cur.next = l1 || l2;
  return dummy.next;
}

// 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 merge_two_lists(
    mut l1: Option<Box<ListNode>>,
    mut l2: Option<Box<ListNode>>
  ) -> Option<Box<ListNode>> {
    match (l1.is_some(), l2.is_some()) {
      (false, false) => None,
      (true, false) => l1,
      (false, true) => l2,
      (true, true) => {
        let mut dummy = Box::new(ListNode::new(0));
        let mut cur = &mut dummy;
        while l1.is_some() && l2.is_some() {
          cur.next = if l1.as_ref().unwrap().val < l2.as_ref().unwrap().val {
            let mut res = l1.take();
            l1 = res.as_mut().unwrap().next.take();
            res
          } else {
            let mut res = l2.take();
            l2 = res.as_mut().unwrap().next.take();
            res
          };
          cur = cur.next.as_mut().unwrap();
        }
        cur.next = if l1.is_some() { l1.take() } else { l2.take() };
        dummy.next.take()
      }
    }
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function (l1, l2) {
  const dummy = new ListNode(0);
  let cur = dummy;
  while (l1 && l2) {
    if (l1.val <= l2.val) {
      cur.next = l1;
      l1 = l1.next;
    } else {
      cur.next = l2;
      l2 = l2.next;
    }
    cur = cur.next;
  }
  cur.next = l1 || l2;
  return dummy.next;
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
    ListNode dummy = new ListNode(0);
    ListNode cur = dummy;
    while (l1 != null && l2 != null) {
      if (l1.val <= l2.val) {
        cur.next = l1;
        l1 = l1.next;
      } else {
        cur.next = l2;
        l2 = l2.next;
      }
      cur = cur.next;
    }
    cur.next = l1 == null ? l2 : l1;
    return dummy.next;
  }
}

方法二：递归

我们先判断 l1 和 l2 中有没有一个为空，如果有一个为空，那么我们直接返回另一个链表即可。

接下来，我们比较 l1 和 l2 的值：

• 如果 l1 的值小于等于 l2 的值，我们递归调用 mergeTwoLists(l1.next, l2)，并将 l1.next 指向返回的链表，然后返回 l1。
• 如果 l1 的值大于 l2 的值，我们递归调用 mergeTwoLists(l1, l2.next)，并将 l2.next 指向返回的链表，然后返回 l2。

时间复杂度 $O(m + n)$，空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别为两个链表的长度。

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, x):
#     self.val = x
#     self.next = None


class Solution:
  def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
    if l1 is None or l2 is None:
      return l1 or l2
    if l1.val <= l2.val:
      l1.next = self.mergeTwoLists(l1.next, l2)
      return l1
    else:
      l2.next = self.mergeTwoLists(l1, l2.next)
      return l2

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode(int x) { val = x; }
 * }
 */
class Solution {
  public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    if (l1 == null) {
      return l2;
    }
    if (l2 == null) {
      return l1;
    }
    if (l1.val <= l2.val) {
      l1.next = mergeTwoLists(l1.next, l2);
      return l1;
    } else {
      l2.next = mergeTwoLists(l1, l2.next);
      return l2;
    }
  }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *   int val;
 *   ListNode *next;
 *   ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
  ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    if (!l1) {
      return l2;
    }
    if (!l2) {
      return l1;
    }
    if (l1->val <= l2->val) {
      l1->next = mergeTwoLists(l1->next, l2);
      return l1;
    } else {
      l2->next = mergeTwoLists(l1, l2->next);
      return l2;
    }
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
  if l1 == nil {
    return l2
  }
  if l2 == nil {
    return l1
  }
  if l1.Val <= l2.Val {
    l1.Next = mergeTwoLists(l1.Next, l2)
    return l1
  } else {
    l2.Next = mergeTwoLists(l1, l2.Next)
    return l2
  }
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function mergeTwoLists(l1: ListNode | null, l2: ListNode | null): ListNode | null {
  if (l1 == null || l2 == null) {
    return l1 || l2;
  }
  if (l1.val < l2.val) {
    l1.next = mergeTwoLists(l1.next, l2);
    return l1;
  }
  l2.next = mergeTwoLists(l1, l2.next);
  return l2;
}

// 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 merge_two_lists(
    l1: Option<Box<ListNode>>,
    l2: Option<Box<ListNode>>
  ) -> Option<Box<ListNode>> {
    match (l1, l2) {
      (Some(mut n1), Some(mut n2)) => {
        if n1.val < n2.val {
          n1.next = Self::merge_two_lists(n1.next, Some(n2));
          Some(n1)
        } else {
          n2.next = Self::merge_two_lists(Some(n1), n2.next);
          Some(n2)
        }
      }
      (Some(node), None) => Some(node),
      (None, Some(node)) => Some(node),
      (None, None) => None,
    }
  }
}

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *   this.val = val;
 *   this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var mergeTwoLists = function (l1, l2) {
  if (!(l1 && l2)) {
    return l1 || l2;
  }
  if (l1.val < l2.val) {
    l1.next = mergeTwoLists(l1.next, l2);
    return l1;
  } else {
    l2.next = mergeTwoLists(l2.next, l1);
    return l2;
  }
};

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   public int val;
 *   public ListNode next;
 *   public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
  public ListNode MergeTwoLists(ListNode l1, ListNode l2) {
    if (l1 == null) {
      return l2;
    }
    if (l2 == null) {
      return l1;
    }
    if (l1.val <= l2.val) {
      l1.next = MergeTwoLists(l1.next, l2);
      return l1;
    } else {
      l2.next = MergeTwoLists(l1, l2.next);
      return l2;
    }
  }
}