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feat: update leetcode solutions: No.0445. Add Two Numbers II
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‎README.md

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@@ -75,6 +75,7 @@
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### 链表
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- [两数相加](/solution/0000-0099/0002.Add%20Two%20Numbers/README.md)
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- [两数相加 II](/solution/0400-0499/0445.Add%20Two%20Numbers%20II/README.md)
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- [从尾到头打印链表](/lcof/面试题06.%20从尾到头打印链表/README.md)
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- [删除链表的节点](/lcof/面试题18.%20删除链表的节点/README.md)
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- [删除排序链表中的重复元素](/solution/0000-0099/0083.Remove%20Duplicates%20from%20Sorted%20List/README.md)

‎README_EN.md

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@@ -74,6 +74,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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### Linked List
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- [Add Two Numbers](/solution/0000-0099/0002.Add%20Two%20Numbers/README_EN.md)
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- [Add Two Numbers II](/solution/0400-0499/0445.Add%20Two%20Numbers%20II/README_EN.md)
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- [Delete Node in a Linked List](/solution/0200-0299/0237.Delete%20Node%20in%20a%20Linked%20List/README_EN.md)
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- [Remove Duplicates from Sorted List](/solution/0000-0099/0083.Remove%20Duplicates%20from%20Sorted%20List/README_EN.md)
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- [Remove Duplicates from Sorted List II](/solution/0000-0099/0082.Remove%20Duplicates%20from%20Sorted%20List%20II/README_EN.md)

‎solution/0400-0499/0445.Add Two Numbers II/README.md

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<!-- 这里可写通用的实现逻辑 -->
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利用栈将数字逆序。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
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s1, s2 = [], []
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while l1:
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s1.append(l1.val)
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l1 = l1.next
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while l2:
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s2.append(l2.val)
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l2 = l2.next
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carry, dummy = 0, ListNode(-1)
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while s1 or s2 or carry:
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carry += (0 if not s1 else s1.pop()) + (0 if not s2 else s2.pop())
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# 创建结点，利用头插法将结点插入链表
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node = ListNode(carry % 10)
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node.next = dummy.next
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dummy.next = node
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carry //= 10
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return dummy.next
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
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Deque<Integer> s1 = new ArrayDeque<>();
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Deque<Integer> s2 = new ArrayDeque<>();
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for (; l1 != null; l1 = l1.next) {
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s1.push(l1.val);
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}
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for (; l2 != null; l2 = l2.next) {
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s2.push(l2.val);
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}
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int carry = 0;
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ListNode dummy = new ListNode(-1);
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while (!s1.isEmpty() || !s2.isEmpty() || carry != 0) {
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carry += (s1.isEmpty() ? 0 : s1.pop()) + (s2.isEmpty() ? 0 : s2.pop());
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// 创建结点，利用头插法将结点插入链表
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ListNode node = new ListNode(carry % 10);
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node.next = dummy.next;
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dummy.next = node;
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carry /= 10;
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}
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return dummy.next;
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}
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}
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```
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### **...**

‎solution/0400-0499/0445.Add Two Numbers II/README_EN.md

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@@ -35,13 +35,64 @@ What if you cannot modify the input lists? In other words, reversing the lists i
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### **Python3**
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
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s1, s2 = [], []
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while l1:
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s1.append(l1.val)
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l1 = l1.next
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while l2:
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s2.append(l2.val)
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l2 = l2.next
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carry, dummy = 0, ListNode(-1)
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while s1 or s2 or carry:
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carry += (0 if not s1 else s1.pop()) + (0 if not s2 else s2.pop())
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node = ListNode(carry % 10)
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node.next = dummy.next
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dummy.next = node
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carry //= 10
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return dummy.next
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```
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### **Java**
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
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Deque<Integer> s1 = new ArrayDeque<>();
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Deque<Integer> s2 = new ArrayDeque<>();
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for (; l1 != null; l1 = l1.next) {
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s1.push(l1.val);
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}
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for (; l2 != null; l2 = l2.next) {
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s2.push(l2.val);
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}
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int carry = 0;
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ListNode dummy = new ListNode(-1);
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while (!s1.isEmpty() || !s2.isEmpty() || carry != 0) {
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carry += (s1.isEmpty() ? 0 : s1.pop()) + (s2.isEmpty() ? 0 : s2.pop());
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ListNode node = new ListNode(carry % 10);
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node.next = dummy.next;
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dummy.next = node;
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carry /= 10;
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}
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return dummy.next;
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}
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}
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```
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### **...**

‎solution/0400-0499/0445.Add Two Numbers II/Solution.java

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@@ -1,3 +1,11 @@
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
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Deque<Integer> s1 = new ArrayDeque<>();
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for (; l2 != null; l2 = l2.next) {
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s2.push(l2.val);
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}
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ListNode head = null;
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int carry = 0;
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while (!s1.isEmpty() || !s2.isEmpty() || carry > 0) {
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ListNode dummy = new ListNode(-1);
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while (!s1.isEmpty() || !s2.isEmpty() || carry != 0) {
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carry += (s1.isEmpty() ? 0 : s1.pop()) + (s2.isEmpty() ? 0 : s2.pop());
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ListNode h = new ListNode(carry % 10);
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h.next = head;
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head = h;
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ListNode node = new ListNode(carry % 10);
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node.next = dummy.next;
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dummy.next = node;
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carry /= 10;
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}
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return head;
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return dummy.next;
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}
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}
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}

‎solution/0400-0499/0445.Add Two Numbers II/Solution.py

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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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7+
class Solution:
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def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
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s1, s2 = [], []
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while l1:
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s1.append(l1.val)
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l1 = l1.next
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while l2:
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s2.append(l2.val)
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l2 = l2.next
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carry, dummy = 0, ListNode(-1)
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while s1 or s2 or carry:
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carry += (0 if not s1 else s1.pop()) + (0 if not s2 else s2.pop())
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node = ListNode(carry % 10)
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node.next = dummy.next
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dummy.next = node
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carry //= 10
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return dummy.next

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