177 lines
5.2 KiB
Markdown
177 lines
5.2 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/add-two-numbers/description/
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## 题目描述
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```
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You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
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You may assume the two numbers do not contain any leading zero, except the number 0 itself.
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Example
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Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
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Output: 7 -> 0 -> 8
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Explanation: 342 + 465 = 807.
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```
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## 思路
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设立一个表示进位的变量carried，建立一个新链表，
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把输入的两个链表从头往后同时处理，每两个相加，将结果加上carried后的值作为一个新节点到新链表后面。
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(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)
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## 关键点解析
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1. 链表这种数据结构的特点和使用
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2. 用一个carried变量来实现进位的功能，每次相加之后计算carried，并用于下一位的计算
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## 代码
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* 语言支持：JS，C++
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JavaScript:
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```js
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} l1
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* @param {ListNode} l2
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* @return {ListNode}
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*/
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var addTwoNumbers = function(l1, l2) {
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if (l1 === null || l2 === null) return null
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// 使用dummyHead可以简化对链表的处理，dummyHead.next指向新链表
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let dummyHead = new ListNode(0)
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let cur1 = l1
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let cur2 = l2
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let cur = dummyHead // cur用于计算新链表
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let carry = 0 // 进位标志
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while (cur1 !== null || cur2 !== null) {
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let val1 = cur1 !== null ? cur1.val : 0
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let val2 = cur2 !== null ? cur2.val : 0
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let sum = val1 + val2 + carry
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let newNode = new ListNode(sum % 10) // sum%10取模结果范围为0~9，即为当前节点的值
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carry = sum >= 10 ? 1 : 0 // sum>=10，carry=1，表示有进位
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cur.next = newNode
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cur = cur.next
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if (cur1 !== null) {
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cur1 = cur1.next
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}
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if (cur2 !== null) {
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cur2 = cur2.next
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}
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}
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if (carry > 0) {
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// 如果最后还有进位，新加一个节点
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cur.next = new ListNode(carry)
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}
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return dummyHead.next
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};
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```
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C++
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> C++代码与上面的JavaScript代码略有不同：将carry是否为0的判断放到了while循环中
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```c++
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
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ListNode* ret = nullptr;
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ListNode* cur = nullptr;
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int carry = 0;
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while (l1 != nullptr || l2 != nullptr || carry != 0) {
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carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
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auto temp = new ListNode(carry % 10);
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carry /= 10;
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if (ret == nullptr) {
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ret = temp;
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cur = ret;
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}
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else {
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cur->next = temp;
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cur = cur->next;
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}
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l1 = l1 == nullptr ? nullptr : l1->next;
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l2 = l2 == nullptr ? nullptr : l2->next;
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}
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return ret;
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}
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};
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```
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## 拓展
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通过单链表的定义可以得知，单链表也是递归结构，因此，也可以使用递归的方式来进行reverse操作。
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> 由于单链表是线性的，使用递归方式将导致栈的使用也是线性的，当链表长度达到一定程度时，递归会导致爆栈，因此，现实中并不推荐使用递归方式来操作链表。
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### 描述
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1. 将两个链表的第一个节点值相加，结果转为0-10之间的个位数，并设置进位信息
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2. 将两个链表第一个节点以后的链表做带进位的递归相加
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3. 将第一步得到的头节点的next指向第二步返回的链表
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### C++实现
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```C++
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// 普通递归
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class Solution {
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public:
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ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
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return addTwoNumbers(l1, l2, 0);
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}
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private:
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ListNode* addTwoNumbers(ListNode* l1, ListNode* l2, int carry) {
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if (l1 == nullptr && l2 == nullptr && carry == 0) return nullptr;
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carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
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auto ret = new ListNode(carry % 10);
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ret->next = addTwoNumbers(l1 == nullptr ? l1 : l1->next,
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l2 == nullptr ? l2 : l2->next,
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carry / 10);
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return ret;
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}
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};
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// （类似）尾递归
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class Solution {
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public:
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ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
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ListNode* head = nullptr;
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addTwoNumbers(head, nullptr, l1, l2, 0);
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return head;
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}
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private:
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void addTwoNumbers(ListNode*& head, ListNode* cur, ListNode* l1, ListNode* l2, int carry) {
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if (l1 == nullptr && l2 == nullptr && carry == 0) return;
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carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
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auto temp = new ListNode(carry % 10);
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if (cur == nullptr) {
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head = temp;
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cur = head;
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} else {
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cur->next = temp;
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cur = cur->next;
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}
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addTwoNumbers(head, cur, l1 == nullptr ? l1 : l1->next, l2 == nullptr ? l2 : l2->next, carry / 10);
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}
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};
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```
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