LeetCode 2. Add Two Numbers（單連結清單求和）

題目描述：

    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.

例子：

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

分析：

   題意：給定兩個非空、非負整數元素的單連結清單，它們可以看作兩個整數的逆序排列形式。求兩個整數和的逆序排列單連結清單。假設不存在前置零的問題。

  思路：考察整數加法和單連結清單插入法。我們用指針p，q分别指向兩個單連結清單頭部順序周遊，用指針r開辟新的單連結清單結點空間儲存和。每次相加結果模10作為數值，除以10作為進制，r指針采用尾插法儲存結果（頭插法會形成逆序、尾插法會保持正序），最後傳回r指針作為結果。假設兩個單連結清單結點數為m、n，時間複雜度為O(max(m, n))。

代碼：

#include <bits/stdc++.h>

using namespace std;

// Definition for singly-linked list
struct ListNode{
	int val;
	ListNode *next;
	ListNode(int x): val(x), next(NULL){}
};

class Solution {
private: 
	// Tail insertion method: keep the nodes' order
	void add(ListNode *&r, int digit){
		r->next = new ListNode(digit);
		r = r->next;			
	}

public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        // Exceptional Case: 
		if(!l1){
			return l2;
		}
		if(!l2){
			return l1;
		}
		// Add an extra empty ListNode(-1) is convenient for the operation of pointer r
		ListNode *p = l1, *q = l2, *l3 = new ListNode(-1), *r = l3;
		int digit = 0, carry = 0;
		while(p && q){
			digit = (p->val + q->val + carry) % 10;
			carry = (p->val + q->val + carry) / 10;
			add(r, digit);
			p = p->next;
			q = q->next;
		}
		while(p){
			digit = (p->val + carry) % 10;
			carry = (p->val + carry) / 10;
			add(r, digit);
			p = p->next;
		}
		while(q){
			digit = (q->val + carry) % 10;
			carry = (q->val + carry) / 10;
			add(r, digit);
			q = q->next;
		}
		if(carry){
			add(r, 1);
		}
		return l3->next;
    }
};