21. Merge Two Sorted Lists

Description

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

Constraints:

The number of nodes in both lists is in the range [0, 50].
-100 <= Node.val <= 100
Both list1 and list2 are sorted in non-decreasing order.

Solutions

Solution 1: Recursion

First, we judge whether the linked lists $l_1$ and $l_2$ are empty. If one of them is empty, we return the other linked list. Otherwise, we compare the head nodes of $l_1$ and $l_2$:

If the value of the head node of $l_1$ is less than or equal to the value of the head node of $l_2$, we recursively call the function $mergeTwoLists(l_1.next, l_2)$, and connect the head node of $l_1$ with the returned linked list head node, and return the head node of $l_1$.
Otherwise, we recursively call the function $mergeTwoLists(l_1, l_2.next)$, and connect the head node of $l_2$ with the returned linked list head node, and return the head node of $l_2$.

The time complexity is $O(m + n)$, and the space complexity is $O(m + n)$. Here, $m$ and $n$ are the lengths of the two linked lists respectively.

Python Code

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(
        self, list1: Optional[ListNode], list2: Optional[ListNode]
    ) -> Optional[ListNode]:
        if list1 is None or list2 is None:
            return list1 or list2
        if list1.val <= list2.val:
            list1.next = self.mergeTwoLists(list1.next, list2)
            return list1
        else:
            list2.next = self.mergeTwoLists(list1, list2.next)
            return list2

Java Code

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        if (list1 == null) {
            return list2;
        }
        if (list2 == null) {
            return list1;
        }
        if (list1.val <= list2.val) {
            list1.next = mergeTwoLists(list1.next, list2);
            return list1;
        } else {
            list2.next = mergeTwoLists(list1, list2.next);
            return list2;
        }
    }
}

C++ Code

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
        if (!list1) return list2;
        if (!list2) return list1;
        if (list1->val <= list2->val) {
            list1->next = mergeTwoLists(list1->next, list2);
            return list1;
        } else {
            list2->next = mergeTwoLists(list1, list2->next);
            return list2;
        }
    }
};

Go Code

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(list1 *ListNode, list2 *ListNode) *ListNode {
	if list1 == nil {
		return list2
	}
	if list2 == nil {
		return list1
	}
	if list1.Val <= list2.Val {
		list1.Next = mergeTwoLists(list1.Next, list2)
		return list1
	} else {
		list2.Next = mergeTwoLists(list1, list2.Next)
		return list2
	}
}

TypeScript Code

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/**
 * 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(list1: ListNode | null, list2: ListNode | null): ListNode | null {
    if (list1 == null || list2 == null) {
        return list1 || list2;
    }
    if (list1.val < list2.val) {
        list1.next = mergeTwoLists(list1.next, list2);
        return list1;
    } else {
        list2.next = mergeTwoLists(list1, list2.next);
        return list2;
    }
}

Rust Code

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// 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(
        list1: Option<Box<ListNode>>,
        list2: Option<Box<ListNode>>
    ) -> Option<Box<ListNode>> {
        match (list1, list2) {
            (None, None) => None,
            (Some(list), None) => Some(list),
            (None, Some(list)) => Some(list),
            (Some(mut list1), Some(mut list2)) => {
                if list1.val < list2.val {
                    list1.next = Self::merge_two_lists(list1.next, Some(list2));
                    Some(list1)
                } else {
                    list2.next = Self::merge_two_lists(Some(list1), list2.next);
                    Some(list2)
                }
            }
        }
    }
}

JavaScript Code

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/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} list1
 * @param {ListNode} list2
 * @return {ListNode}
 */
var mergeTwoLists = function (list1, list2) {
    if (!list1 || !list2) {
        return list1 || list2;
    }
    if (list1.val <= list2.val) {
        list1.next = mergeTwoLists(list1.next, list2);
        return list1;
    } else {
        list2.next = mergeTwoLists(list1, list2.next);
        return list2;
    }
};

C# Code

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode MergeTwoLists(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode();
        ListNode cur = dummy;
        while (list1 != null && list2 != null)
        {
            if (list1.val <= list2.val)
            {
                cur.next = list1;
                list1 = list1.next;
            }
            else
            {
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        cur.next = list1 == null ? list2 : list1;
        return dummy.next;
    }
}

Ruby Code

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# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} list1
# @param {ListNode} list2
# @return {ListNode}
def merge_two_lists(list1, list2)
    dummy = ListNode.new()
    cur = dummy
    while list1 && list2
        if list1.val <= list2.val
            cur.next = list1
            list1 = list1.next
        else
            cur.next = list2
            list2 = list2.next
        end
        cur = cur.next
    end
    cur.next = list1 || list2
    dummy.next
end

Solution 2: Iteration

We can also use iteration to implement the merging of two sorted linked lists.

First, we define a dummy head node $dummy$, then loop through the two linked lists, compare the head nodes of the two linked lists, add the smaller node to the end of $dummy$, until one of the linked lists is empty, then add the remaining part of the other linked list to the end of $dummy$.

Finally, return $dummy.next$.

The time complexity is $O(m + n)$, where $m$ and $n$ are the lengths of the two linked lists respectively. Ignoring the space consumption of the answer linked list, the space complexity is $O(1)$.