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

初次看题没懂。

要点是 each of their nodes contain a single digit.

上面的例子是：

2+5=7；

4+6=10；10不满足 a  single digit条件，0保存到当前结点中，进位1就要保存到下一个结点。

思路：

思路非常简单，分别遍历两个链表，并且用一个变量表示是否有进位。某个链表遍历结束之后再将另一个链表连接在结果链表之后即可，若最后有进位需要添加一位。

给出一种解答：Runtime: 420 ms

head 相当于头指针，方便遍历结果链表

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode result = new ListNode(0);        ListNode head = result;        int sum = 0;        int carry = 0;//jin wei        while(l1 != null || l2 != null) {            int val1 = 0;            if(l1 != null){                val1 = l1.val;                l1 = l1.next;            }            int val2 = 0;            if(l2 != null) {                val2 = l2.val;                l2 = l2.next;            }                        sum = val1 + val2 + carry;            result.next = new ListNode(sum%10);            result = result.next;            carry = sum/10;                                }        if(carry == 1){            result.next = new ListNode(1);        }        return head.next;    }}

 

 

 九章算法网给出的答案是：

Runtime: 540 ms

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        if(l1 == null && l2 == null) {            return null;        }                    ListNode head = new ListNode(0);        ListNode point = head;        int carry = 0;        while(l1 != null && l2!=null){            int sum = carry + l1.val + l2.val;            point.next = new ListNode(sum % 10);            carry = sum / 10;            l1 = l1.next;            l2 = l2.next;            point = point.next;        }                while(l1 != null) {            int sum =  carry + l1.val;            point.next = new ListNode(sum % 10);            carry = sum /10;            l1 = l1.next;            point = point.next;        }                while(l2 != null) {            int sum =  carry + l2.val;            point.next = new ListNode(sum % 10);            carry = sum /10;            l2 = l2.next;            point = point.next;        }                if (carry != 0) {            point.next = new ListNode(carry);        }        return head.next;    }}