Leetcode# 2. Add Two Numbers（链表模拟大数算法）

Add Two Numbers（链表）

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.

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

题意

给定两个非空的链表，表示两个非负整数。数字以相反的顺序存储，每一个节点包含一个数字。将两个数相加并将结果作为链表返回。
注意返回的结果的列表也是相反的。
运算过程：

2 4 3
5 6 4
—————
7 0 8

特别像字符串求大数算法，这里题目已经把字符串翻转，简单了不少。

解决方法一：C++版

思路很简单，麻烦之处在于指针的使用。sum指向头结点,temp代表移动的指针,在末尾添加元素的操作为：
temp ->next = 当前数
指针后移：
temp ->next = next

/** * 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) {

        int carry = 0;//进位
        ListNode *sum = NULL,*temp = NULL;

        while(l1 != NULL || l2 != NULL || carry != 0)
        {
            int now = 0;//当前位数
            if(l1!=NULL)
            {
                now += l1 -> val;
                l1 = l1 -> next;
            }
            if(l2!=NULL)
            {
                now += l2 -> val;
                l2 = l2 -> next;
            }
            if(carry != 0)
                now += carry;

            carry = now / 10; 
            now = now % 10;
            //将int型的now转换为ListNode注意看上面的构造函数要传参数的
            ListNode *p = new ListNode(now);

            //头指针指向temp
            if(sum == NULL)
            {
                temp = p;
                sum  = temp;
            }
            else 
            {
                temp -> next = p;
                temp = temp -> next;
            }
        }
        return sum;
    }
};

解决方法二：Python版

解法思路同上。

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self,x):
# self.val = x
# self.next = None

class Solution(object):
    def addTwoNumbers(self,l1,l2):
        """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """
        sum = None
        temp = None
        carry = 0
        while l1 != None or l2 != None or carry != 0:
            now = 0
            if l1 != None:
                now = now + l1.val
                l1 = l1.next
            if l2 != None:
                now = now + l2.val
                l2 = l2.next
            if carry != 0:
                now = now + carry
            carry = now / 10
            now = now % 10
            q = ListNode(now)
            if sum == None:
                temp = q
                sum = temp
            else:
                temp.next = q
                temp = temp.next

        return sum

总结

这两种语言思路是一样的，如果不太懂大数算法的可以在这里学习一下：http://www.voidcn.com/article/p-bgqfrziw-ke.html

本地github代码：https://github.com/xuna123/LeetCode/blob/master/Leetcode%23%202.%20Add%20Two%20Numbers.md