LeetCode.2(445) Add Two numbers(II)

题目1(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.

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

分析1:

/** * 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) {        //采用新建链表保存结果        ListNode pre=new ListNode(0);        ListNode head=pre;        int remain=0;        while(l1!=null||l2!=null||remain!=0){            //建立一个新节点            ListNode cur=new ListNode(0);            int sum;            sum=(l1==null?0:l1.val)+(l2==null?0:l2.val)+remain;            //保存节点的值            cur.val=sum%10;            remain=sum/10;                        //pre节点指向当前节点            pre.next=cur;            pre=pre.next;                        l1=l1==null?l1:l1.next;            l2=l2==null?l2:l2.next;        }        return head.next;    }}

题目2(Add Two Numbers II):

You are given two non-empty linked lists representing two non-negative integers. 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

分析2:

/** * 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) {        //第2题是往后进位，而该题是往前进位        //给定两个非空0链表,使用stack实现        Stack<Integer> s1=new Stack<Integer>();        Stack<Integer> s2=new Stack<Integer>();                //入栈        while(l1!=null){            s1.push(l1.val);            l1=l1.next;        }        while(l2!=null){            s2.push(l2.val);            l2=l2.next;        }                int sum=0;        ListNode dummy=new ListNode(0);        while(!s1.empty()||!s2.empty()){            if(!s1.empty())sum+=s1.pop();            if(!s2.empty())sum+=s2.pop();            dummy.val=sum%10;            //如果超过10，则将十位作为head节点。            ListNode head=new ListNode(sum/10);            head.next=dummy;            dummy=head;            //进位            sum/=10;        }        //判断是否头节点是否是0，若不是则表明head未有进位，否则直接取head。        return dummy.val==0?dummy.next:dummy;    }}