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目录
Description:
Add Two Numbers
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.
Solution:
/**
* 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) {
int carry=0;//进位缓存
ListNode dummyHead = new ListNode(0);
ListNode curr = dummyHead;
while(l1!=null||l2!=null){
int a1=(l1!=null)?l1.val:0;
int a2=(l2!=null)?l2.val:0;
int sum =a1+a2+carry;
carry = sum/10;
curr.next = new ListNode(sum%10);
curr=curr.next;
if(l1!=null) l1=l1.next;
if(l2!=null) l2=l2.next;
}
if(carry>0) curr.next = new ListNode(carry);
return dummyHead.next;
}
}
Submission:
Summary:
注意测试,第一次提交忽略了当l1或l2为空时,直接加和会有空指针错误。第二次是因为没有测试最后进位,如5+5=10,最后carry为1,但是由于已经结束循环,故需在后面加上最后进位。