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题目地址

https://leetcode.com/problems/add-two-numbers/description/

题目描述

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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

思路

设立一个表示进位的变量carried，建立一个新链表， 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。

2.addTwoNumbers

(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)

关键点解析

  1. 链表这种数据结构的特点和使用

  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

代码

  • 语言支持：JS，C++

JavaScript:

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function(l1, l2) {
  var carried = 0; // 用于进位
  const head = new ListNode();
  const noop = {
    val: 0,
    next: null
  };
  let currentL1 = l1;
  let currentL2 = l2;
  let currentNode = head; // 返回的链表的当前node
  let newNode; // 声明在外面节省内存
  let previousNode; // 记录前一个节点，便于删除最后一个节点

  while (currentL1 || currentL2) {
    newNode = new ListNode(0);
    
    currentNode.val =
      ((currentL1 || noop).val + (currentL2 || noop).val + carried) % 10;

    currentNode.next = newNode;
    previousNode = currentNode;
    currentNode = newNode;

    if ((currentL1 || noop).val + (currentL2 || noop).val + carried >= 10) {
      carried = 1;
    } else {
      carried = 0;
    }

    currentL1 = (currentL1 || noop).next;
    currentL2 = (currentL2 || noop).next;
  }

  if (carried) {
    // 还有位没进呢
    previousNode.next = new ListNode(carried)
  } else {
    previousNode.next = null;
  }

  return head;
};

C++

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* ret = nullptr;
        ListNode* cur = nullptr;
        int carry = 0;
        while (l1 != nullptr || l2 != nullptr || carry != 0) {
            carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
            auto temp = new ListNode(carry % 10);
            carry /= 10;
            if (ret == nullptr) {
                ret = temp;
                cur = ret;
            }
            else {
                cur->next = temp;
                cur = cur->next;
            }
            l1 = l1 == nullptr ? nullptr : l1->next;
            l2 = l2 == nullptr ? nullptr : l2->next;
        }
        return ret;
    }
};

拓展

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。

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。

描述

  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
  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
  3. 将第一步得到的头节点的next指向第二步返回的链表

C++实现

// 普通递归
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        return addTwoNumbers(l1, l2, 0);
    }
    
private:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2, int carry) {
        if (l1 == nullptr && l2 == nullptr && carry == 0) return nullptr;
        carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
        auto ret = new ListNode(carry % 10);
        ret->next = addTwoNumbers(l1 == nullptr ? l1 : l1->next,
                                 l2 == nullptr ? l2 : l2->next,
                                 carry / 10);
        return ret;
    }
};
// （类似）尾递归
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = nullptr;
        addTwoNumbers(head, nullptr, l1, l2, 0);
        return head;
    }
    
private:
    void addTwoNumbers(ListNode*& head, ListNode* cur, ListNode* l1, ListNode* l2, int carry) {
        if (l1 == nullptr && l2 == nullptr && carry == 0) return;
        carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
        auto temp = new ListNode(carry % 10);
        if (cur == nullptr) {
            head = temp;
            cur = head;
        } else {
            cur->next = temp;
            cur = cur->next;
        }
        addTwoNumbers(head, cur, l1 == nullptr ? l1 : l1->next, l2 == nullptr ? l2 : l2->next, carry / 10);
    }
};