1669. 合并两个链表

題解：
模拟链表操作
/**
 * 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) {
        int i = 0;
        ListNode head = new ListNode(-1, list1);
        ListNode index = list1;
        ListNode right = null;
        ListNode left = null;
        // 拿到删除始节点的前一个节点
        while (i < a - 1 && index != null) {
            index = index.next;
            i++;
        }
        left = index;
        if (a == 0) head.next = list2;
        // 拿到删除末节点的后一个节点
        while (i < b + 1 && index != null) {
            index = index.next;
            i++;
        }
        right = index;
        // 拼接
        left.next = list2;
        while (list2.next != null) {
            list2 = list2.next;
        }
        list2.next = right;
        // 返回开头
        return head.next;
    }
}
原文地址：https://www.cnblogs.com/eiffelzero/p/17077116.html