1669. Merge In Between Linked Lists

SourceBiweekly Contest 40 Q2DifficultyMediumRating1428

Description

You are given two linked lists: list1 and list2 of sizes n and m respectively.

Remove list1's nodes from the a^th node to the b^th node, and put list2 in their place.

The blue edges and nodes in the following figure indicate the result:

Build the result list and return its head.

Example 1:

Input: list1 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
Output: [10,1,13,1000000,1000001,1000002,5]
Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.

Example 2:

Input: list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
Output: [0,1,1000000,1000001,1000002,1000003,1000004,6]
Explanation: The blue edges and nodes in the above figure indicate the result.

Constraints:

3 <= list1.length <= 10⁴
1 <= a <= b < list1.length - 1
1 <= list2.length <= 10⁴

Solutions

Solution 1: Simulation

Thinking

Replace indices \([a,b]\) of \(\textit{list1}\) with \(\textit{list2}\). Find the predecessor of \(a\) and node \(b\), link the predecessor to \(\textit{list2}\)'s head, then \(\textit{list2}\)'s tail to \(b\)'s successor.

Two pointers walk the needed steps, a third scan finds \(\textit{list2}\)'s end, and \(b.\textit{next}\) is cleared.

We can directly simulate the operations described in the problem.

In the implementation, we use two pointers \(p\) and \(q\), both initially pointing to the head node of \(\textit{list1}\).

Then we move pointers \(p\) and \(q\) forward, until pointer \(p\) points to the node before the \(a\)-th node in \(\textit{list1}\), and pointer \(q\) points to the \(b\)-th node in \(\textit{list1}\). At this point, we set the \(\textit{next}\) pointer of \(p\) to the head node of \(\textit{list2}\), and set the \(\textit{next}\) pointer of the tail node of \(\textit{list2}\) to the node pointed to by the \(\textit{next}\) pointer of \(q\). This completes the operation required by the problem.

The time complexity is \(O(m + n)\), and the space complexity is \(O(1)\), where \(m\) and \(n\) are the lengths of \(\textit{list1}\) and \(\textit{list2}\) respectively.

Python3JavaC++GoTypeScriptC#

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeInBetween(
        self, list1: ListNode, a: int, b: int, list2: ListNode
    ) -> ListNode:
        p = q = list1
        for _ in range(a - 1):
            p = p.next
        for _ in range(b):
            q = q.next
        p.next = list2
        while p.next:
            p = p.next
        p.next = q.next
        q.next = None
        return list1

/**
 * 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 mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
        ListNode p = list1, q = list1;
        while (--a > 0) {
            p = p.next;
        }
        while (b-- > 0) {
            q = q.next;
        }
        p.next = list2;
        while (p.next != null) {
            p = p.next;
        }
        p.next = q.next;
        q.next = null;
        return list1;
    }
}

/**
 * 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* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {
        auto p = list1, q = list1;
        while (--a) {
            p = p->next;
        }
        while (b--) {
            q = q->next;
        }
        p->next = list2;
        while (p->next) {
            p = p->next;
        }
        p->next = q->next;
        q->next = nullptr;
        return list1;
    }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
    p, q := list1, list1
    for ; a > 1; a-- {
        p = p.Next
    }
    for ; b > 0; b-- {
        q = q.Next
    }
    p.Next = list2
    for p.Next != nil {
        p = p.Next
    }
    p.Next = q.Next
    q.Next = nil
    return list1
}

/**
 * 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 mergeInBetween(
    list1: ListNode | null,
    a: number,
    b: number,
    list2: ListNode | null,
): ListNode | null {
    let p = list1;
    let q = list1;
    while (--a > 0) {
        p = p.next;
    }
    while (b-- > 0) {
        q = q.next;
    }
    p.next = list2;
    while (p.next) {
        p = p.next;
    }
    p.next = q.next;
    q.next = null;
    return list1;
}

/**
 * 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 MergeInBetween(ListNode list1, int a, int b, ListNode list2) {
        ListNode p = list1, q = list1;
        while (--a > 0) {
            p = p.next;
        }
        while (b-- > 0) {
            q = q.next;
        }
        p.next = list2;
        while (p.next != null) {
            p = p.next;
        }
        p.next = q.next;
        q.next = null;
        return list1;
    }
}

1669. Merge In Between Linked Lists

Description

Solutions

Solution 1: Simulation

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