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
题意:
给你两个表示两个非负数字的链表。数字以相反的顺序存储,其节点包含单个数字,将这两个数字相加并将其作为一个链表返回。
解法:
需要注意的点:
1)两个list可能不一样长
2)两个digits 相加如果大于等于10,需要进位
3)输出的node要reverse(反向)
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
题意:
给你两个表示两个非负数字的链表。数字以相反的顺序存储,其节点包含单个数字,将这两个数字相加并将其作为一个链表返回。
解法:
需要注意的点:
1)两个list可能不一样长
2)两个digits 相加如果大于等于10,需要进位
3)输出的node要reverse(反向)
实现:
public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { int carry = 0; ListNode newNode = new ListNode(0); ListNode p1 = l1; ListNode p2 = l2; ListNode p3 = newNode; while(null != p1 || null != p2) { if(null != p1) { carry += p1.val; p1 = p1.next; } if(null != p2) { carry += p2.val; p2 = p2.next; } p3.next = new ListNode(carry%10); p3 = p3.next; carry /= 10; } if(carry == 1) p3.next = new ListNode(1); return newNode.next; } }