问题描述:
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
解题思路:
设立一个头ListNode和尾ListNode,两个List分别从头开始,两两相加并且设立进位carry,如果相加超过10则carry为1,否则为零。同时需要考虑一个List还有长度,另一个已经结束的情况。
代码如下:
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ListNode head = new ListNode(0); ListNode tail = head; int sum = 0; int carry = 0; while(l1 != null || l2 != null){ if(l1 == null){ sum = l2.val + carry; l2 = l2.next; } else if (l2 == null){ sum = l1.val + carry; l1 = l1.next; } else{ sum = l1.val + l2.val + carry; l1 = l1.next; l2 = l2.next; } if(sum >= 10){ carry = sum / 10; sum = sum % 10; } else{ carry = 0; } tail.next = new ListNode(sum); tail = tail.next; } if(carry != 0){ tail.next = new ListNode(carry); tail = tail.next; } return head.next; } }