题目描述
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.
样例
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
思路分析
题意即,两链表节点分别相加,考虑两链表不为空的情况下,处理进位情况,分别相加即可,其余见注释。
代码
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode sentinel = new ListNode(0);//创建新链表 ListNode d = sentinel; int sum = 0; while (l1!=null || l2!=null) { sum /=10; //sum依次与进位,l1,l2相加 if (l1 != null) { sum += l1.val; l1 = l1.next; } if (l2 != null) { sum += l2.val; l2 = l2.next; } d.next = new ListNode(sum % 10); //d的下一节点为两数之和,注意%是保证个位数 d = d.next; //删除d节点,用d.next替代 } if (sum/10 == 1)//考虑最后两位和大于10的情况 { d.next = new ListNode(1); //因为无后续节点,不用删除了 } return sentinel.next; }