class Solution {
public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
ListNode p1 = list1; // 找 a 的前驱
for (int i = 0; i < a - 1; i++) {
p1 = p1.next;
}
ListNode p2 = p1.next; // p2 必定不为 null,找 b 的后继
for (int i = 0; i < b - a + 1; i++) {
p2 = p2.next;
}
ListNode p3 = list2;
while (p3.next != null) {
p3 = p3.next;
}
p1.next = list2;
p3.next = p2;
return list1;
}
}