leetcode-2.Add Two Numbers

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本文介绍了解决LeetCode上第2题“两数相加”的方法。题目要求将两个非空链表表示的非负整数相加，并以链表形式返回结果。文章提供了详细的C++实现代码，利用了STL中的div_t结构来处理进位问题。

leetcode-2.Add Two Numbers

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.

本题需要考虑的就是两个节点数字相加若大于10需要进位的问题，可以用到STL中的div_t结构，该结构含两个成员，分别是 quot(quotient，商) 与 rem(remainder，余数). 可以将rem作为相加后节点的值，而将quot作为下一个节点的进位。

sloution.h

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x):val(x),next(NULL){}
};

class Solution
{
public:
    ListNode * addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        ListNode header(0), *tail = &header;
        for (div_t sum{ 0,0 }; sum.quot || l1 || l2; tail = tail->next)
        {
            if (l1) { sum.quot += l1->val; l1 = l1->next; }
            if (l2) { sum.quot += l2->val; l2 = l2->next; }
            sum = div(sum.quot, 10);
            tail->next= new ListNode(sum.rem);
        }
        return header.next;
    }

};

测试

#include<iostream>
#include"solution.h"
using namespace std;

ListNode *create_linkedlist(std::initializer_list<int> lst)
{
    auto iter = lst.begin();
    ListNode *head = lst.size() ? new ListNode(*iter++) : NULL;
    for (ListNode *cur = head; iter != lst.end(); cur = cur->next)
        cur->next = new ListNode(*iter++);
    return head;
}

int main()
{
    Solution s;
    ListNode *l1 = create_linkedlist({ 2,4,3 });
    ListNode *l2 = create_linkedlist({ 5,6,4 });
    ListNode *ret = s.addTwoNumbers(l1, l2);
    for (ListNode *cur = ret; cur; cur = cur->next)
        cout << cur->val << "->";
    cout << "\b\b  " << endl;
    return 0;
}