2. Add Two Numbers

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

思路：每一位相加，记住进位 （X/10）和当前位（X%10）

/**
 * 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) {
        int carry = 0;
        ListNode head = new ListNode(0);
        ListNode p1 = l1, p2=l2, p3=head;
        while(p1!=null||p2!=null){
            if (p1!=null){
                carry += p1.val;
                p1 = p1.next;
            }

            if (p2!=null){
                carry += p2.val;
                p2 = p2.next;
            }

            int currect = carry%10;
            ListNode nextNode = new ListNode(currect);
            p3.next = nextNode;
            p3 = p3.next;
            carry = carry/10;
        }

        if (carry==1){
            ListNode nextNodeOne = new ListNode(1);
            p3.next = nextNodeOne;
        }

        return head.next;
    }
}

思路2：同样的思路，写的更容易理解

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        int sum = 0;
        int remain = 0;
        ListNode head = new ListNode(0);
        ListNode curr = head;

        while (l1!=null||l2!=null){
             if (l1!=null&&l2!=null){
                sum = l1.val+l2.val+carry;
                carry = sum/10;
                remain = sum%10;
                curr.next = new ListNode(remain);
                curr = curr.next;
                l1 = l1.next;
                l2 = l2.next;
            }else if (l1!=null&&l2==null){
                sum = l1.val+carry;
                carry = sum/10;
                remain = sum%10;
                curr.next = new ListNode(remain);
                curr = curr.next;
                l1 = l1.next;
             }else{
                sum = l2.val+carry;
                carry = sum/10;
                remain = sum%10;
                curr.next = new ListNode(remain);
                curr = curr.next;
                l2 = l2.next;
             }
        }

        if (carry>0){
                curr.next = new ListNode(carry);
                curr = curr.next;
        }

        return head.next;

    }
}

C++ code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry = 0;
        int sum = 0;
        int remain = 0;
        ListNode* newHeader = new ListNode(-1); 
        ListNode* curr = newHeader; 

        while(l1 || l2){
            if (l1 && l2){
                sum = l1->val+l2->val+carry;
                carry = sum/10;
                remain = sum%10;
                curr->next = new ListNode(remain);
                curr = curr->next;
                l1 = l1->next;
                l2 = l2->next;
            }else if(l1){
                sum = l1->val+carry;
                carry = sum/10;
                remain = sum%10;
                curr->next = new ListNode(remain);
                curr = curr->next;
                l1 = l1->next;
            }else if(l2){
                sum = l2->val+carry;
                carry = sum/10;
                remain = sum%10;
                curr->next = new ListNode(remain);
                curr = curr->next;
                l2 = l2->next;
            }            
        }

        if (carry>0){
            curr->next = new ListNode(carry);
        }

        return newHeader->next;   
    }
};