题目:
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.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
思路:
1)由于两个链表的长度不一定相同,而且有可能有空,所以首先应该先排除空链表情况
2)我习惯解决问题是由特例到一般,即先考虑最特殊的情况,再推出最普遍一般的情况。
a. 两个链表都只有一个节点时,只需计算和,判断值是否>10 ,是则进位,否则以该值创建单节点链表即可。
b. 当两个链表都不只一个节点则需要考虑双方后续节点是否为null,在这里我将判断是否null 和进位融合在一起,当判断前一位相加的和>10 后,若next不为空则next.val++。
c. 迭代addTwoNumbers(l1.next,l2.next)。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1 == null) l1 = new ListNode(0);
if(l2 == null) l2 = new ListNode(0);
if(l1.next == null && l2.next == null){
int val = l1.val + l2.val;
if(val > 9){
int geWei = val%10;
ListNode node = new ListNode(geWei);
int shiWei = val/10;
node.next = new ListNode(shiWei);
return node;
}else
return new ListNode(val);
}else{
int val = l1.val + l2.val;
if(val > 9){
val = val - 10;
if(l1.next != null) l1.next.val++;
else if (l2.next != null) l2.next.val++;
}
ListNode node = new ListNode (val);
node.next = addTwoNumbers(l1.next,l2.next);
return node;
}
}
}