0445. 两数相加 II【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - 栈

1. 📝 题目描述

• leetcode

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

示例 1：

输入：l1 = [7,2,4,3], l2 = [5,6,4]
输出：[7,8,0,7]

示例 2：

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[8,0,7]

示例 3：

输入：l1 = [0], l2 = [0]
输出：[0]

---

提示：

• 链表的长度范围为 [1, 100]
• 0 <= node.val <= 9
• 输入数据保证链表代表的数字无前导 0

进阶：如果输入链表不能翻转该如何解决？

2. 🎯 s.1 - 栈

c

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    int s1[100], s2[100], t1 = 0, t2 = 0;
    while (l1) { s1[t1++] = l1->val; l1 = l1->next; }
    while (l2) { s2[t2++] = l2->val; l2 = l2->next; }
    int carry = 0;
    struct ListNode* head = NULL;
    while (t1 > 0 || t2 > 0 || carry) {
        int sum = carry;
        if (t1 > 0) sum += s1[--t1];
        if (t2 > 0) sum += s2[--t2];
        carry = sum / 10;
        struct ListNode* node = (struct ListNode*)malloc(sizeof(struct ListNode));
        node->val = sum % 10;
        node->next = head;
        head = node;
    }
    return head;
}

js

/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
  const s1 = [],
    s2 = []
  while (l1) {
    s1.push(l1.val)
    l1 = l1.next
  }
  while (l2) {
    s2.push(l2.val)
    l2 = l2.next
  }
  let carry = 0,
    head = null
  while (s1.length || s2.length || carry) {
    const sum = (s1.pop() || 0) + (s2.pop() || 0) + carry
    carry = Math.floor(sum / 10)
    const node = new ListNode(sum % 10)
    node.next = head
    head = node
  }
  return head
}

py

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        s1, s2 = [], []
        while l1:
            s1.append(l1.val)
            l1 = l1.next
        while l2:
            s2.append(l2.val)
            l2 = l2.next
        carry, head = 0, None
        while s1 or s2 or carry:
            total = (s1.pop() if s1 else 0) + (s2.pop() if s2 else 0) + carry
            carry = total // 10
            node = ListNode(total % 10)
            node.next = head
            head = node
        return head

• 时间复杂度：$O(max(m,n))$
• 空间复杂度：$O(m+n)$

算法思路：

• 将两个链表分别压入栈，从低位开始相加
• 用头插法构建结果链表，处理进位