Add Two Numbers
- Total Accepted: 160702
- Total Submissions: 664770
- Difficulty: Medium
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
本题的思路其实很简单，两个链表的结构都是从低位到高位的顺序，税后要求返回的链表也是这样的结构。所以我们只需要依次循环两个链表，将对应位的数相加，并判断是否有进位，有进位则将进位累加到下一位即可。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Num2 {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if(l1 == null){
return l2 ;
}
if(l2 == null){
return l1 ;
}
int nextBit = (l1.val + l2.val)/10 ;
int curBit = (l1.val + l2.val)%10 ;
ListNode head = new ListNode(curBit) ;
ListNode temp = head ;
l1 = l1.next ;
l2 = l2.next ;
//当l1、l2对应位均存在时，进行计算
while(l1 != null && l2 != null){
curBit = (l1.val + l2.val + nextBit) % 10 ;
nextBit = (l1.val + l2.val + nextBit)/10 ;
ListNode node = new ListNode(curBit) ;
temp.next = node ;
temp = temp.next ;
l1 = l1.next ;
l2 = l2.next ;
}
//判断l1是否结束，没有结束继续
while(l1 != null){
curBit = (l1.val + nextBit) % 10 ;
nextBit = (l1.val + nextBit)/10 ;
ListNode node = new ListNode(curBit) ;
temp.next = node ;
temp = temp.next ;
l1 = l1.next ;
}
//判断l2是否结束，没有结束继续
while(l2 != null){
curBit = (l2.val + nextBit) % 10 ;
nextBit = (l2.val + nextBit)/10 ;
ListNode node = new ListNode(curBit) ;
temp.next = node ;
temp = temp.next ;
l2 = l2.next ;
}
//判断最后的进位位是否为0 ，不为0则需要保存下一位
if(nextBit != 0){
ListNode node = new ListNode(nextBit) ;
temp.next = node ;
}
return head ;
}
}
