1669. Merge In Between Linked Lists

⭐️⭐️

# 題目敘述

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

Remove list1 's nodes from the ath node to the bth 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.

// 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;

    }

}

# 解題思路

# Solution

class Solution {

    public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {

        ListNode head = new ListNode();

        head.next = list1;

        ListNode prev = head;

        for (int i = 0; i < a; i++) {

            prev = prev.next;

        }

        ListNode next = prev;

        for (int i = 0; i < b - a + 1; i++) {

            next = next.next;

        }

        prev.next = list2;

        while (list2.next != null) {

            list2 = list2.next;

        }

        list2.next = next.next;

        return head.next;

    }

}