LeetCode002

Add two numbers

Posted by coding-zen on October 31, 2018

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.
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode c1 = l1;
        ListNode c2 = l2;
        //定义一个链表作为链表头
        ListNode head = new ListNode(0);
        //指向head
        ListNode d = head;
        int sum = 0;
        while (c1 != null || c2 != null) {
            sum /= 10;
            if (c1 != null) {
                sum += c1.val;
                c1 = c1.next;
            }
            if (c2 != null) {
                sum += c2.val;
                c2 = c2.next;
            }
            d.next = new ListNode(sum % 10);
            d = d.next;
        }
        if (sum / 10 == 1)
            d.next = new ListNode(1);
        return head.next;
    }    

}


 public ListNode addTwoNumbers(ListNode l1,ListNode l2){
        ListNode dummyHead = new ListNode(0);
        ListNode p=l1,q=l2,curr=dummyHead;
        int carry=0;//进位
        while(p!=null || q!=null){
            int x=(p!=null)?p.val:0;
            int y=(q!=null)?q.val:0;
            int sum=x+y+carry;
            carry=sum/10;
            curr.next=new ListNode(sum%10);
            curr=curr.next;
            if(p!=null) p=p.next;
            if(q!=null) q=q.next;
        }
        if(carry>0){
            curr.next=new ListNode(carry);
        }
        
        return dummyHead.next;
}

public class ListNode
{
    int val;
    ListNode next;
    
    public ListNode(int x){
        val=x;
    }
}