链表求和

 链表求和

 

 

 描述
 笔记

 数据

 评测

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中

相反

的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

您在真实的面试中是否遇到过这个题？ 

Yes

样例

给出两个链表 

3->1->5->null

 和 

5->9->2->null

，返回

8->0->8->null

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
/**
* @param l1: the first list
* @param l2: the second list
* @return: the sum list of l1 and l2
*/
public ListNode addLists(ListNode l1, ListNode l2) {
// write your code here
ListNode root=null,cur=null;
int flag = 0;
while(l1 !=null && l2!=null){
int val = l1.val + l2.val +flag;
flag = val / 10;
val = val % 10;

ListNode temp = new ListNode(val);
if(root == null){
root = temp;
}
if(cur != null){
cur.next = temp;
}
cur = temp;
l1 = l1.next;
l2 = l2.next;

}
while(l1 != null){
int val = l1.val + flag;
flag = val / 10;
val %= 10;
ListNode temp = new ListNode(val);
if(root == null){
root = temp;
}
if(cur != null){
cur.next = temp;
}
cur = temp;
l1 = l1.next;
}
while(l2 != null){
int val = l2.val + flag;
flag = val / 10;
val %= 10;
ListNode temp = new ListNode(val);
if(root == null){
root = temp;
}
if(cur != null){
cur.next = temp;
}
cur = temp;
l2 = l2.next;
}
if(flag == 1){
ListNode temp = new ListNode(flag);
cur.next = temp;
}
return root;
}
}