[LeetCode]Add Two Numbers II

Question

You are given two linked lists representing two non-negative numbers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

本题难度Medium。

stack法

【复杂度】

时间 O(N) 空间 O(N)

【思路】

本题与两个Array相加那题非常相似，但是链表可不是能从后往前读取的，那么就用stack。

【注意】

我本来想在某个input链表上进行修改的（这样是可以的，如果超出了链表头就new ListNode），不过为保持代码干净就不这么写（这样也会对input产生side effect，在工程应用会对client造成一定干扰，不利于函数式编程）。

【代码】

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
//require
ListNode ans=null;
Stack<ListNode> stack1=new Stack<ListNode>(),stack2=new Stack<ListNode>();
//invariant
while(l1!=null||l2!=null){
if(l1!=null){
stack1.push(l1);
l1=l1.next;
}
if(l2!=null){
stack2.push(l2);
l2=l2.next;
}
}
int a=0,b=0,carry=0,sum=0;
while(!stack1.isEmpty()||!stack2.isEmpty()){
if(!stack1.isEmpty())
a=stack1.pop().val;
else
a=0;
if(!stack2.isEmpty())
b=stack2.pop().val;
else
b=0;
sum=a+b+carry;
ListNode tmp=new ListNode(sum%10);
tmp.next=ans;
ans=tmp;
carry=sum/10;
}
if(carry!=0){
ListNode tmp=new ListNode(carry);
tmp.next=ans;
ans=tmp;
}

//ensure
return ans;
}
}