[Amazon] Add Two Numbers I(II)

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

思路：本质上还是ListNode的题，但是又加上两数相加，需要考虑借位，所以是interview好题

"模10除10"

/** * 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) {        if(l1==null || l2==null){            return null;        }                ListNode dummy=new ListNode(-1);        ListNode head=dummy;        int carry=0;        while(l1!=null && l2!=null){            int sum=carry+l1.val+l2.val;            head.next=new ListNode(sum%10);            carry=sum/10;            head=head.next;            l1=l1.next;            l2=l2.next;        }                while(l1!=null){                                 //l2短，只剩l1            int sum=carry+l1.val;            head.next=new ListNode(sum%10);            carry=sum/10;            head=head.next;            l1=l1.next;        }        while(l2!=null){                                 //l1短，只剩l2            int sum=carry+l2.val;            head.next=new ListNode(sum%10);            carry=sum/10;            head=head.next;            l2=l2.next;        }        if(carry!=0){                                    //如果最后一位借位之后不是一位数字了            head.next=new ListNode(carry%10);        }        return dummy.next;        }}

Follow up:

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

/** * 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 addLists2(ListNode l1, ListNode l2) {        if(l1==null && l2==null){            return null;        }                ListNode n1=reverse(l1);                                   //反转        ListNode n2=reverse(l2);                                   //反转        ListNode dummy=new ListNode(-1);        ListNode head=dummy;        int carry=0;        while(n1!=null && n2!=null){            int sum=carry+n1.val+n2.val;            head.next=new ListNode(sum%10);            carry=sum/10;            head=head.next;            n1=n1.next;            n2=n2.next;        }        while(n1!=null){            int sum=carry+n1.val;            head.next=new ListNode(sum%10);            carry=sum/10;            head=head.next;            n1=n1.next;        }        while(n2!=null){            int sum=carry+n2.val;            head.next=new ListNode(sum%10);            carry=sum/10;            head=head.next;            n2=n2.next;        }        if(carry!=0){            head.next=new ListNode(carry%10);        }        return reverse(dummy.next);                                //反转    }         private ListNode reverse(ListNode head){        ListNode pre=null;        while(head!=null){            ListNode temp=head.next;            head.next=pre;            pre=head;            head=temp;        }        return pre;    }}