LeetCode2. Add Two Numbers

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

使用链表，要记录链表的表头：用一个 ListNode 记录链表头，用一个 ListNode 记录当前节点。
像这种加法要考虑 3 种情况：
1、两个链表的长度不一样；
2、其中一个链表为空；
3、类似 5 + 5 = 10，这种溢出一位的情况；

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {           ListNode result = new ListNode(0);    ListNode curr = result;    int c = 0;    while(l1 != null || l2 != null){        int i1 = l1 != null ? l1.val : 0;        int i2 = l2 != null ? l2.val : 0;        ListNode node = new ListNode((i1 + i2 + c) % 10);        c = (i1 + i2 + c) / 10;        curr.next = node;        curr = curr.next;        l1 = l1 != null ? l1.next : l1;        l2 = l2 != null ? l2.next : l2;    }    if(c == 0){                  return result.next;    }    else{        curr.next = new ListNode(1);        return result.next;    }}

public ListNode addTwoNumbers2(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;}

public ListNode addTwoNumbers3(ListNode l1, ListNode l2) {    ListNode c1 = l1;    ListNode c2 = l2;    ListNode sentinel = new ListNode(0);    ListNode d = sentinel;    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 sentinel.next;}