LeetCode Ex02 Add Two Numbers

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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

题目

该题给出两个链表，以逆序存储，返回其相加后的结果。其实只要把两个链表依次相加就行了，它所说的逆序只是储存的形式，结果是正序的。例如（3->4->2）+（2->5->6）
结果应该是（5->9->8）而不是（8->9>5），返回的（5->9->8）储存的形式是895所以是正确答案。

建立一个新链表，然后把输入的两个链表从头往后遍历，每两个相加，其中若后面的节点为Null，则默认为0，然后添加一个新节点到新链表后面，注意进位，其中carry维护进位，若两个节点相加大于10，则carry=1，在下一次两个节点相加时还要加上这个carry，同时carry也会重置。注意：最高位的进位在循环结束后还要特殊处理。

    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;
    }

下面是稍微修改较简介的版本

    public ListNode addTwoNumbers2(ListNode l1, ListNode l2) {
        ListNode preHead = new ListNode(0);
        ListNode runner = preHead;
        int carry = 0;
        while (l1 != null || l2 != null || carry != 0) {
            if (l1 != null) {
                carry += l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                carry += l2.val;
                l2 = l2.next;
            }
            runner.next = new ListNode(carry % 10);
            runner = runner.next;
            carry /= 10;
        }
        return preHead.next;
    }