leetcode题解-2. Add Two Numbers

题意：给你两个表示两个非负数字的链表。数字以相反的顺序存储，其节点包含单个数字。将这两个数字相加并将其作为一个链表返回。

例子：输入: (2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 0 -> 8
解释：342 + 465 = 807.

分析：这道题比较简单，两个链表相加的问题，思路很明确，就是按照位数读下去，维护当前位和进位，时间复杂度是O(n),空间复杂度是O(1).

另外本题是链表部分的第一题，所以也附上链表部分的主函数和打印函数，以后的链表部分题目为了简介不在包含主函数。值得注意的是，因为本题定义了链表节点的数据结构，因此包含Solution.java和ListNode.java两部分的代码。以后为了精简博客，ListNode.java部分的代码也不再附上。

Solution.java

class Solution {   public  ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if(l1 == null && l2 == null) return null;        ListNode point = new ListNode(0);        ListNode head = point;        int carry = 0;        while(l1 != null && l2 != null){            int sum = carry + l1.val + l2.val;            point.next = new ListNode(sum % 10);            point = point.next;            carry = sum /10;            l1 = l1.next;            l2 = l2.next;        }        while(l1 != null){            int sum = carry + l1.val;            point.next = new ListNode(sum % 10);            point = point.next;            carry = sum /10;            l1 = l1.next;        }        while(l2 != null){            int sum = carry + l2.val;            point.next = new ListNode(sum % 10);            point = point.next;            carry = sum /10;            l2 = l2.next;        }        if(carry != 0){            point.next = new ListNode(carry);        }        return head.next;    }    public static void print(ListNode head){        while(head != null){            System.out.println(head.val);            head = head.next;        }    }    public static void main(String[] args) {        ListNode l1 = new ListNode(2);        ListNode l2 = new ListNode(4);        ListNode l3 = new ListNode(3);        ListNode l4 = new ListNode(5);        ListNode l5 = new ListNode(6);        ListNode l6 = new ListNode(4);        l1.next = l2;        l2.next = l3;        l4.next = l5;        l5.next = l6;        print(addTwoNumbers(l1, l4));    }}

ListNode.java

// Definition for singly-linked list.     public class ListNode {        int val;        ListNode next;        ListNode(int x) { val = x; }     }