算法分析与设计——LeetCode:2.Add Two Numbers

题目

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

/**
* 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) {
};

思路

　　每个链表都表示一个多位数，每个节点里的val是该位的数字，头节点是个位，我们要将其每一位相加和进位。一开始我试图将两个链表的整数分别读取出来并相加，但是发现当数字位数特别多时会超出int和long long int的范围，所以只能老实从头节点开始，每一位分别进行相加和进位，并将结果连接成一个新链表。如下代码，用n来进位，当相加结果为两位数，n赋值为1。

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) {
ListNode *p1, *head, *p2, *p;
int n = 0;
p1 = l1;
p2 = l2;
head = new ListNode(0);
p = head;
while (p1 != NULL || p2 != NULL || n != 0) {
if (p1 != NULL) {
n += p1->val;
p1 = p1->next;
}
if (p2 != NULL) {
n += p2->val;
p2 = p2->next;
}
p->next = new ListNode(n%10);
p = p->next;
if (n >= 10) {
n = 1;
} else {
n = 0;
}
}
return head->next;
}
};