Add Two Numbers

LeetCode #2

Description:

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.

Example:

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

Idea:

用carry表示进位，要处理好到最后是否创建新node的情况。

Method 1 更简洁

Code:

Method 1

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int carry=0;
        ListNode dummy(-1);
        ListNode *curr=&dummy;

        while(l1!=NULL || l2!=NULL){
            int sum=carry;
            if(l1!=NULL){
                sum += l1->val;
                l1=l1->next;
            }
            if(l2!=NULL){
                sum += l2->val;
                l2=l2->next;
            }
            curr->next=new ListNode(sum%10);
            curr=curr->next;
            carry=sum/10;
        }
        if(carry){
            curr->next = new ListNode(1);
        }
        return dummy.next;
    }
};

Method 2

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        if(l1==NULL || l2==NULL) return l1==NULL? l2 : l1;

        ListNode dummy(-1);
        ListNode *curr = &dummy;
        int carry = 0;
        int add_digit;

        while(l1!=NULL && l2!=NULL){
            add_digit = l1->val + l2->val + carry;
            carry = add_digit/10;
            curr->next = new ListNode(add_digit%10);
            curr=curr->next;

            l1=l1->next;
            l2=l2->next;
        }

        ListNode *l_remain = l1==NULL? l2: l1;
        while(l_remain!=NULL && carry!=0){
            add_digit = l_remain->val + carry;
            carry = add_digit/10;
            curr->next = new ListNode(add_digit%10);
            curr=curr->next;
            l_remain = l_remain->next;
        }

        if(carry==0)
            curr->next=l_remain;
        else
            curr->next=new ListNode(carry);

        return dummy.next;

    }
};