leetcode - 2.Add Two Numbers-优快云博客

本文介绍了一种解决两数相加问题的方法，通过链表形式存储逆序的整数，并提供两种解决方案。一种是创建一个新的链表来存储结果，另一种是在原有链表上进行调整。

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

Solution1(better):

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode head = new ListNode(0);
    ListNode l = head;
    int sum = 0;

    while (l1 != null || l2 != null) {
        sum /= 10;

        if (l1 != null) {
            sum += l1.val;
            l1 = l1.next;
        }
        if (l2 != null) {
            sum += l2.val;
            l2 = l2.next;
        }

        l.next = new ListNode(sum % 10);
        l = l.next;
    }

    if(sum / 10 > 0){
        l.next = new ListNode(sum / 10);
    }

    return head.next;
}

Solution2:

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode l = l1;
    ListNode zero = new ListNode(0);

    while (l1 != null) {
        l1.val += l2.val;

        adjust(l1);

        if (l1.next == null) {
            l1.next = l2.next;
            l2.next = zero;
        }

        if (l2.next == null) {
            l2 = zero;
        } else {
            l2 = l2.next;
        }

        l1 = l1.next;
    }

    return l;
}

private void adjust(ListNode l) {
    if (l.val % 10 == 0) {
        if (l.next == null) {
            l.next = new ListNode(0);
        }

        l.next.val += 1;
        l.val -= 10;
    }
}