『1』迭代法
class Solution {
// Iteration
// N is the size of l1, M is the size of l2
// Time Complexity: O(max(M, N))
// Space Complexity: O(max(M, N)) if dummy is counted else O(1)
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int next1 = 0;
int total = 0;
ListNode dummy = new ListNode(); // 虚拟头节点
ListNode cur = dummy;
while (l1 != null && l2 != null) {
total = l1.val + l2.val + next1;
cur.next = new ListNode(total % 10);
next1 = total / 10;
l1 = l1.next;
l2 = l2.next;
cur = cur.next;
}
while (l1 != null) {
total = l1.val + next1;
cur.next = new ListNode(total % 10);
next1 = total / 10;
l1 = l1.next;
cur = cur.next;
}
while (l2 != null) {
total = l2.val + next1;
cur.next = new ListNode(total % 10);
next1 = total / 10;
l2 = l2.next;
cur = cur.next;
}
// 处理最后的进位
if (next1 != 0) {
cur.next = new ListNode(next1);
}
return dummy.next;
}
}
扩展：哨兵节点通常被称为"哑结点"或"虚拟节点"。它们的作用是充当链表中的一个假头，其目的是简化链表操作的边界条件处理。它们通常用于链表的插入、删除和遍历操作中，可以帮助简化代码逻辑，避免对空链表或头结点进行特殊处理。在一些算法和数据结构中，哨兵节点能够简化代码的实现，提高程序的健壮性和可读性。
『2』递归法
class Solution {
// Recursion
// N is the size of l1, M is the size of l2
// Time Complexity: O(max(M, N))
// Space Complexity: O(max(M, N)) if dummy is counted else O(1)
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int total = l1.val + l2.val;
int next1 = total / 10;
ListNode res = new ListNode(total % 10);
if (l1.next != null || l2.next != null || next1 != 0) {
l1 = l1.next != null ? l1.next : new ListNode(0);
l2 = l2.next != null ? l2.next : new ListNode(0);
l1.val += next1;
res.next = addTwoNumbers(l1, l2);
}
return res;
}
}