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
題意：
給你兩個(gè)表示兩個(gè)非負(fù)數(shù)字的鏈表。數(shù)字以相反的順序存儲(chǔ)，其節(jié)點(diǎn)包含單個(gè)數(shù)字，將這兩個(gè)數(shù)字相加并將其作為一個(gè)鏈表返回。
解法：
需要注意的點(diǎn)：
1）兩個(gè)list可能不一樣長(zhǎng)
2）兩個(gè)digits 相加如果大于等于10，需要進(jìn)位
3）輸出的node要reverse（反向）
實(shí)現(xiàn)：
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;
ListNode newNode = new ListNode(0);
ListNode p1 = l1;
ListNode p2 = l2;
ListNode p3 = newNode;
while(null != p1 || null != p2)
{
if(null != p1)
{
carry += p1.val;
p1 = p1.next;
}
if(null != p2)
{
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);
p3 = p3.next;
carry /= 10;
}
if(carry == 1)
p3.next = new ListNode(1);
return newNode.next;
}
}