2. Add Two Numbers

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本文介绍了一种解决两个链表表示的非负整数相加问题的方法。通过逆序存储每个数字，并使用进位标志来处理加法运算，最终将结果返回为一个新的链表。

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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

思路:就跟普通的加法一样，设置一个进位标志carry就行

代码如下(已通过leetcode)

public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode result = new ListNode(0);
ListNode currentNode = result;
ListNode pre = null;
int carry = 0;
while(l1 != null && l2 != null){
currentNode.val = (l1.val + l2.val + carry) % 10;
carry = (l1.val + l2.val + carry) / 10;
pre = currentNode;
currentNode = new ListNode(0);
pre.next = currentNode;
l1 = l1.next;
l2 = l2.next;
}
currentNode = addWithCarry(l1, l2, carry);
pre.next = currentNode;
return result;

}
public ListNode addWithCarry(ListNode l1, ListNode l2, int carry){
if (l1 == null && l2 == null) {
if (carry != 0) {
ListNode cur = new ListNode(0);
cur.val = carry;
return cur;
}
return null;
}
if (l1 == null) {
return addWithCarry(l2, l1, carry);
}
ListNode result = new ListNode(0);
ListNode cur = result;
ListNode pre = null;
while (l1 != null) {
cur.val = (l1.val + carry) % 10;
carry = (l1.val + carry) / 10;
pre = cur;
cur = new ListNode(carry);
pre.next = cur;
l1 = l1.next;
}
if (pre.next.val == 0) {
pre.next = null;
}
return result;
}
}