标签：des style class blog code ext
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
Solution: link list
简单，注意两点：
1. 加到最后，可能还有进位
2. 链表实现时，可以多申请一个node的空间为x，返回x.next，可使代码更简洁。
#include <iostream>
#include <vector>
// 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 out = ListNode(0);
ListNode *tail = &out;
if(!l1 || !l2)
{
return NULL;
}
int carry = 0;
while(l1 && l2)
{
int s = carry + l1->val + l2->val;
carry = s / 10;
s = s % 10;
l1 = l1->next;
l2 = l2->next;
ListNode *temp = new ListNode(s);
tail->next = temp;
tail = tail->next;
}
if(l2)
{
l1 = l2;
}
while(l1)
{
int s = l1->val + carry;
carry = s / 10;
s = s % 10;
l1 = l1->next;
ListNode *temp = new ListNode(s);
tail->next = temp;
tail = tail->next;
}
if(carry)
{
ListNode *temp = new ListNode(carry);
tail->next = temp;
tail = tail->next;
}
return out.next;
}
};
[leetcode] 2. Add Two Numbers,布布扣,bubuko.com
标签：des style class blog code ext
原文地址：http://www.cnblogs.com/xjsxjtu/p/3793600.html
