2. Add Two Numbers

官方的链接:2. Add Two Numbers

Description :

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.


问题描述

给定2个非空的代表非负整数的链表,数字是倒过来存储的,用链表表示求和。可以假设没有前导数字0。

思路

使用迭代解法,注意最后的进位。

[github-here]

 public class Q2_AddTwoNumbers {

     /**
* 迭代解法
*
* @param ListNode
* l1
* @param ListNode
* l2
* @return
*/
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (null == l1 && null == l2) {
return new ListNode(0);
}
// 保存链表头,便于创建结点和最后返回结果
ListNode headNode = new ListNode(0);
ListNode sumNode = headNode;
// 和以及进位
int sum = 0;
int carry = 0;
while (null != l1 && null != l2) {
sum = l1.val + l2.val + carry;
sumNode.next = new ListNode(sum % 10);
carry = sum / 10;
sumNode = sumNode.next;
l1 = l1.next;
l2 = l2.next;
}
while (null != l1) {
sum = l1.val + carry;
sumNode.next = new ListNode(sum % 10);
carry = sum / 10;
sumNode = sumNode.next;
l1 = l1.next;
}
while (null != l2) {
sum = l2.val + carry;
sumNode.next = new ListNode(sum % 10);
carry = sum / 10;
sumNode = sumNode.next;
l2 = l2.next;
}
if (carry > 0) {
sumNode.next = new ListNode(carry);
sumNode = sumNode.next;
}
return headNode.next;
}
}

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