Add Two Numbers

ListNode.java

package Add;

/**
 * Created by chenpeng on 2017/9/4.
 *
 * @version 1.0
 */

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

public class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
    }
}

class Solution {

    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);

        ListNode l2 = new ListNode(9);
        l2.next = new ListNode(9);

        ListNode listNode = addTwoNumbers(l1, l2);
        System.out.println(listNode);
    }

    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode listNode = result;
        int i = 0;
        while (l1 != null || l2 != null) {
            ListNode next = new ListNode(0);
            int temp = 0;
            if (l1 != null) {
                temp += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                temp += l2.val;
                l2 = l2.next;
            }
            int val = temp % 10 + i;
            next.val = val % 10;
            if (val > 9 || temp > 9) {
                i = 1;
            } else {
                i = 0;
            }
            listNode.next = next;
            listNode = listNode.next;
        }
        if (i > 0) {
            listNode.next = new ListNode(1);
        }
        return result.next;
    }
}

Solution.java

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

1. ?表达式的运用，无需if判空，这个想到了，但是没有具体的实现方案，运用不熟悉
2. 命名的方式：carry、sum、p、q、current等
3. 除法直接算出carry
4. next其实对应答案的current
5. curr.next = new ListNode(carry); 这种写法更为规范