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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
链表的对应节点加法。
/**
* 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 *ans = new ListNode(0);
ListNode *pos1 = l1;
ListNode *pos2 = l2;
int num = pos1->val + pos2->val, carry = 0;
if(num > 9)
{
carry = 1;
num -= 10;
}
ans->val = num;
ans->next = nullptr;
pos1 = pos1->next;
pos2 = pos2->next;
ListNode *before = ans;
while(pos1 && pos2)
{
ListNode *tmp = new ListNode(0);
auto num = carry + pos1->val + pos2->val;
carry = 0;
if(num > 9)
{
carry = 1;
num -= 10;
}
tmp->val = num;
before->next = tmp;
before = tmp;
pos1 = pos1->next;
pos2 = pos2->next;
}
while(pos1)
{
ListNode *tmp = new ListNode(0);
auto num = carry + pos1->val;
carry = 0;
if(num > 9)
{
carry = 1;
num -= 10;
}
tmp->val = num;
tmp->next = nullptr;
before->next = tmp;
before = tmp;
pos1 = pos1->next;
}
while(pos2)
{
ListNode *tmp = new ListNode(0);
auto num = carry + pos2->val;
carry = 0;
if(num > 9)
{
carry = 1;
num -= 10;
}
tmp->val = num;
tmp->next = nullptr;
before->next = tmp;
before = tmp;
pos2 = pos2->next;
}
if(carry)
{
ListNode *tmp = new ListNode(carry);
before->next = tmp;
before = tmp;
}
return ans;
}
};
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原文地址:http://my.oschina.net/u/347565/blog/416435