问题描述
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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
解法
复杂度为 O(n^2).
/**
* 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) {
int carry = 0;
ListNode dummyHead(0);
ListNode *curr = &dummyHead, *p= l1, *q=l2;
while(p!=NULL || q!=NULL)
{
int x= (p!=NULL)? p->val: 0;
int y= (q!=NULL)? q->val: 0;
int sum = x+y+carry;
carry=sum/10;
curr->next=new ListNode(sum%10);
curr=curr->next;
if(p!=NULL) p=p->next;
if(q!=NULL) q=q->next;
}
if (carry>0)
{
curr->next=new ListNode(carry);
}
return dummyHead.next;
}
};
使用Python
class Solution(object):
def addTwoNumbers(self, l1, l2):
"""
:type l1: ListNode
:type l2: ListNode
:rtype: ListNode
"""
result = ListNode(0)
p_result = result
p1 = l1
p2 = l2
step_val=0
while p1!=None or p2!=None:
if p1!=None:
val1=p1.val
else:
val1=0
if p2!=None:
val2=p2.val
else:
val2=0
sum_val = val1+val2+step_val
p_result.next=ListNode(sum_val%10)
p_result= p_result.next
step_val=sum_val/10
if p1!=None:
p1 = p1.next
if p2!=None:
p2 = p2.next
if step_val!=0:
p_result.next=ListNode(step_val)
return result.next