LeetCode-445. Add Two Numbers II (JAVA)链表数字加法 leetcode

445. Add Two Numbers II
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7

链表加减，不能反转链表，只能借助栈或者其他容器实现，
【LeetCode-445. Add Two Numbers II (JAVA)链表数字加法】

public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if (l1 == null) return l2; if (l2 == null) return l1; Stack s1 = new Stack<>(); Stack s2 = new Stack<>(); while (l1 != null) { s1.push(l1.val); l1 = l1.next; } while (l2 != null) { s2.push(l2.val); l2 = l2.next; } int sum = 0; // head的下一个结点为curNode ListNode curNode = new ListNode(0); while (!s1.isEmpty() || !s2.isEmpty()) { if (!s1.isEmpty()) sum += s1.pop(); if (!s2.isEmpty()) sum += s2.pop(); // head.val存储进制位，head.val可能为0 ListNode head = new ListNode(sum / 10); // curNode存储结果 curNode.val = sum % 10; head.next = curNode; // curNode往前移动，指向head curNode = head; // 此时sum存储的是进制位 // 下次计算需要用到 sum /= 10; } // 前导0的情况, // curNode为head的引用，可能为0 if (curNode.val == 0) curNode = curNode.next; return curNode; // 前导0的情况 // return curNode.val == 0 ? // curNode.next : curNode; }