Merge Two Sorted Lists

Problem 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.

Examples

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]

Jump to Solution: Python Java C++ JavaScript C#

Python Solution


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

Java Solution


/**
 * 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) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                current.next = list1;
                list1 = list1.next;
            } else {
                current.next = list2;
                list2 = list2.next;
            }
            current = current.next;
        }
        
        current.next = list1 != null ? list1 : list2;
        return dummy.next;
    }
}

C++ Solution


/**
 * 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) {
        ListNode* dummy = new ListNode(0);
        ListNode* current = dummy;
        
        while (list1 && list2) {
            if (list1->val <= list2->val) {
                current->next = list1;
                list1 = list1->next;
            } else {
                current->next = list2;
                list2 = list2->next;
            }
            current = current->next;
        }
        
        current->next = list1 ? list1 : list2;
        
        ListNode* result = dummy->next;
        delete dummy;
        return result;
    }
};

JavaScript Solution


/**
 * 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) {
    const dummy = new ListNode(0);
    let current = dummy;
    
    while (list1 && list2) {
        if (list1.val <= list2.val) {
            current.next = list1;
            list1 = list1.next;
        } else {
            current.next = list2;
            list2 = list2.next;
        }
        current = current.next;
    }
    
    current.next = list1 || list2;
    return dummy.next;
};

C# Solution


/**
 * 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(0);
        ListNode current = dummy;
        
        while (list1 != null && list2 != null) {
            if (list1.val <= list2.val) {
                current.next = list1;
                list1 = list1.next;
            } else {
                current.next = list2;
                list2 = list2.next;
            }
            current = current.next;
        }
        
        current.next = list1 != null ? list1 : list2;
        return dummy.next;
    }
}

Complexity Analysis

Time Complexity: O(n + m) where n and m are the lengths of the input lists
Space Complexity: O(1) as we only use a constant amount of extra space

Solution Explanation

This solution uses an iterative approach to merge two sorted lists. Here's how it works:

Create a dummy node to handle edge cases and simplify the code
Use a current pointer to build the merged list
Compare nodes from both lists:
- Take the smaller value and add it to the merged list
- Move the pointer of the list we took the node from
After one list is exhausted, append the remaining nodes from the other list

Key points:

Dummy node simplifies handling the head of the merged list
The solution maintains the sorted order of the input lists
No extra space is needed as we reuse the existing nodes
The solution handles edge cases (empty lists) properly