LeetCode 2 Add Two Numbers（链表操作）

题目来源：https://leetcode.com/problems/add-two-numbers/

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

题目要求：给出两个链表l1和l2,将相对应的元素相加,如果产生进位则保留个位,相应的十位加在后面的对应元素上.

解体思路：

首先判断链表是否为空,若l1空,直接返回l2,相反返回l2

然后找到l1和l2链表最短的一个长度min_len,进入循环

循环过程中,为了减少空间的开销,可以将链表l1作为最终的结果链表,只需要改变l1的最大容量为l1+l2即可,另外相加过程中需要一个整型temp变量来保存进位.

循环结束之后,判断l1或l2空否(之前计算最短链表时可以做标记),然后将不空的链表中其他元素全部放入结果链表中即可.

提交代码：

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
//        ListNode *pResult = NULL;
//        ListNode **pCur = &pResult;

        ListNode rootNode(0);
        ListNode *pCurNode = &rootNode;
        int a = 0;
        while (l1 || l2)
        {
            int v1 = (l1 ? l1->val : 0);
            int v2 = (l2 ? l2->val : 0);
            int temp = v1 + v2 + a;
            a = temp / 10;
            ListNode *pNode = new ListNode((temp % 10));
            pCurNode->next = pNode;
            pCurNode = pNode;
            if (l1)
                l1 = l1->next;
            if (l2)
                l2 = l2->next;
        }
        if (a > 0)
        {
            ListNode *pNode = new ListNode(a);
            pCurNode->next = pNode;
        }
        return rootNode.next;
    }
};

下面给出数组解决的代码,具体过程相同,只是数据存储结构不同：

#include <bits/stdc++.h>
#define MAX 1000010

using namespace std;

int main()
{
    int n1,n2;
    while(~scanf("%d %d",&n1,&n2))
    {
        int *a=new int[n1+n2+1];
        int *b=new int[n2];
        for(int i=0;i<n1;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<n2;i++)
            scanf("%d",&b[i]);
        int minlen=min(n1,n2);
        int maxlen=n1+n2-minlen;
        int temp=0;
        for(int i=0;i<minlen;i++)
        {
            a[i]=a[i]+b[i]+temp;
            if(a[i]>=10)
            {
                temp=a[i]/10;
                a[i]=a[i]%10;
            }
        }
        for(int j=minlen;j<maxlen;j++)
        {
                if(maxlen==n1)
                    a[j]=a[j];
                else
                    a[j]=b[j];
        }
        for(int i=0;i<maxlen;i++)
            printf("%d ",a[i]);
        printf("
");
    }
}