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

知識點：

本題考查的是單鏈表的基本操作。關(guān)于單鏈表，一些基本操作包括單鏈表的創(chuàng)建，添加，插入，刪除。本題只用到了單鏈表的創(chuàng)建和添加。

首先是結(jié)構(gòu)體的建立，這道題已經(jīng)幫我們建立好了結(jié)構(gòu)體，如下：

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

一個值，一個指向下一節(jié)點的指針，以及一個構(gòu)造函數(shù)。重點注意：這個構(gòu)造函數(shù)中的參數(shù)并未指定缺省值，所以我們在代碼中不能出現(xiàn)這種語句：
<code> q = new ListNode;</code>
而必須要指定參數(shù)值，例如指定參數(shù)值為0，則代碼應(yīng)寫為
<code> q = new ListNode(0);</code>
關(guān)于單鏈表的創(chuàng)建和添加，必須要有3個指針：頭指針head用于指向單鏈表的開頭，指針p指向單鏈表末尾，指針q用來new一塊空間。添加操作就是用q指針new一塊空間，然后讓p指向q，最后p=p->next即可，代碼如下：

ListNode *head, *p, *q;
head = new ListNode;
p = head;
//assume that the singly-linked list is 0->1->2
for (int i=0; i<3; ++i){
    q = new ListNode(i);
    p->next = q;
    p = p->next;
}

解題思路：

本題實際上就是大數(shù)加法。用一個變量carry表示進位，然后對應(yīng)的位數(shù)相加的時候要加上進位，這個和記為oneSum。那么carry更新為oneSum / 10。除此之外，要考慮兩個單鏈表長度不相等的情況。最后注意一下carry值是否有剩余。若有，則在結(jié)果最后添加上carry；若沒有，則結(jié)束。
C++代碼如下：

/**
 * 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* rst = new ListNode(0);
        ListNode *p = rst, *q;
        int carry = 0, oneSum;
        while (l1 != NULL && l2 != NULL){
            oneSum = l1->val + l2->val + carry;
            q = new ListNode(oneSum % 10);
            p->next = q;
            p = p->next;
            carry = oneSum / 10;
            l1 = l1->next;
            l2 = l2->next;
        }
        //to find the remaining number
        ListNode* rmn;
        if (l1 == NULL && l2 == NULL)
            rmn = l1;
        else
            rmn = l1 != NULL? l1:l2;
        //to process the remaining number
        while (rmn != NULL){
            oneSum = rmn->val + carry;
            q = new ListNode(oneSum % 10);
            p->next = q;
            p = p->next;
            carry = oneSum / 10;
            rmn = rmn->next;
        }
        //to check whether carry is still non-zero
        if (carry != 0){
            q = new ListNode(carry);
            p->next = q;
            p = p->next;
        }
        return rst->next;
    }
};