445. 两数相加 II
Difficulty: 中等
给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。
你可以假设除了数字 0 之外，这两个数字都不会以零开头。
进阶：
如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。
示例：
输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7
Solution
Language: ****
/**
* 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) {
//937 + 63 = 1000
//从后往前加
if(l1 == null) return l2;
if(l2 == null) return l1;
ListNode current = l1;
int sum1 = 0;
while(current != null)
{
sum1 = sum1 * 10 + current.val;
current = current.next;
}
current = l2;
int sum2 = 0;
while(current != null)
{
sum2 = sum2 * 10 + current.val;
current = current.next;
}
int sum = sum1 + sum2;
List<Integer> list = new ArrayList();
while(sum != 0)
{
int tmp = sum % 10 ;
list.add(tmp);
sum = sum / 10;
}
//然后构造一颗新链表
ListNode dummy = new ListNode(0);
if(list.size() == 0)
{
return dummy;
}
ListNode first = dummy;
for(int i = list.size() - 1; i >= 0; i--)
{
first.next = new ListNode(list.get(i));
first = first.next;
}
first.next = null;
return dummy.next;
}
}
- 这种方式是去把两个链表的值换成int，然后相加，
输入： [3,9,9,9,9,9,9,9,9,9] [7] 输出： [-2,-9,-4,-9,-6,-7,-2,-9,0] 预期： [4,0,0,0,0,0,0,0,0,6]
- 这种都是会越界的； 所以这种方式是不可取的；
借助ArrayList或者Stack，我们可以实现链表的倒序相加
/**
* 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) {
//937 + 63 = 1000
//从后往前加
if(l1 == null && l2 == null) return null;
List<Integer> list1 = new ArrayList();
ListNode current = l1;
while(current != null)
{
list1.add(current.val);
current = current.next;
}
current = l2;
List<Integer> list2 = new ArrayList();
while(current != null)
{
list2.add(current.val);
current = current.next;
}
int i = list1.size() - 1;
int j = list2.size() - 1;
int carry = 0;
ListNode head = null;
while(i >= 0 || j >= 0 || carry !=0)
{
int s1 = i < 0 ? 0 : list1.get(i);
int s2 = j < 0 ? 0 : list2.get(j);
int sum = s1 + s2 + carry;
ListNode node = new ListNode(sum % 10);
node.next = head;
head = node;
carry = sum / 10;
i--;
j--;
}
return head;
}
}
