12.2 Basic Operations on Linked Lists

206. Reverse Linked List

Problem Description

Reverse a linked list.

Input and Output Example

Input a linked list, output the reversed linked list.

Input: 1->2->3->4->5->nullptr
Output: 5->4->3->2->1->nullptr

Solution Explanation

Reversing a linked list is a fundamental skill that must be mastered. We provide two approaches—recursive and iterative. It's recommended to learn both.

Recursive Approach:

C++

ListNode* reverseList(ListNode* head, ListNode* head_prev = nullptr) {
    if (head == nullptr) {
        return head_prev;
    }
    ListNode* head_next = head->next;
    head->next = head_prev;
    return reverseList(head_next, head);
}

Python

def reverseList(
    head: Optional[ListNode], head_prev: Optional[ListNode] = None
) -> Optional[ListNode]:
    if head is None:
        return head_prev
    head_next = head.next
    head.next = head_prev
    return reverseList(head_next, head)

The non-recursive implementation is as follows:

C++

ListNode* reverseList(ListNode* head) {
    ListNode *head_prev = nullptr, *head_next;
    while (head) {
        head_next = head->next;
        head->next = head_prev;
        head_prev = head;
        head = head_next;
    }
    return head_prev;
}

Python

def reverseList(head: Optional[ListNode]) -> Optional[ListNode]:
    head_prev = None
    while head is not None:
        head_next = head.next
        head.next = head_prev
        head_prev = head
        head = head_next
    return head_prev

21. Merge Two Sorted Lists

Problem Description

Given two sorted linked lists, merge them into one sorted linked list.

Input and Output Example

Input: Two linked lists.
Output: A linked list representing the merged result.

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Solution Explanation

We provide both recursive and non-recursive implementations.
The recursive implementation is as follows:

C++

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
    if (l2 == nullptr) {
        return l1;
    }
    if (l1 == nullptr) {
        return l2;
    }
    if (l1->val < l2->val) {
        l1->next = mergeTwoLists(l1->next, l2);
        return l1;
    }
    l2->next = mergeTwoLists(l1, l2->next);
    return l2;
}

Python

def mergeTwoLists(
    l1: Optional[ListNode], l2: Optional[ListNode]
) -> Optional[ListNode]:
    if l1 is None or l2 is None:
        return l1 or l2
    if l1.val < l2.val:
        l1.next = mergeTwoLists(l1.next, l2)
        return l1
    l2.next = mergeTwoLists(l1, l2.next)
    return l2

The non-recursive implementation is as follows:

C++

ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
    ListNode *dummy = new ListNode(0), *node = dummy;
    while (l1 && l2) {
        if (l1->val < l2->val) {
            node->next = l1;
            l1 = l1->next;
        } else {
            node->next = l2;
            l2 = l2->next;
        }
        node = node->next;
    }
    node->next = l1 == nullptr ? l2 : l1;
    return dummy->next;
}

Python

def mergeTwoLists(
    l1: Optional[ListNode], l2: Optional[ListNode]
) -> Optional[ListNode]:
    dummy = ListNode()
    head = dummy
    
    while l1 and l2:
        if l1.val < l2.val:
            dummy.next = l1
            l1 = l1.next
        else:
            dummy.next = l2
            l2 = l2.next
        dummy = dummy.next
    
    dummy.next = l1 or l2
    return head.next

24. Swap Nodes in Pairs

Problem Description

Given a linked list, swap every two adjacent nodes.

Input and Output Example

Input a linked list, output the linked list after swapping nodes in pairs.

Input: 1->2->3->4
Output: 2->1->4->3

Solution Explanation

Use pointers to perform the swap operation. The problem is not very difficult, but requires careful attention.

C++

ListNode* swapPairs(ListNode* head) {
    ListNode *node1 = head, *node2;
    if (node1 && node1->next) {
        node2 = node1->next;
        node1->next = node2->next;
        node2->next = node1;
        head = node2;
        while (node1->next && node1->next->next) {
            node2 = node1->next->next;
            node1->next->next = node2->next;
            node2->next = node1->next;
            node1->next = node2;
            node1 = node2->next;
        }
    }
    return head;
}

Python

def swapPairs(head: Optional[ListNode]) -> Optional[ListNode]:
    node1 = head
    if node1 is not None and node1.next is not None:
        node2 = node1.next
        node1.next = node2.next
        node2.next = node1
        head = node2
        while node1.next is not None and node1.next.next is not None:
            node2 = node1.next.next
            node1.next.next = node2.next
            node2.next = node1.next
            node1.next = node2
            node1 = node2.next
    return head