Add Two Numbers

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

http://www.lintcode.com/en/problem/add-two-numbers/

Solution

比较简单，注意进位就行了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    ListNode *addLists(ListNode *l1, ListNode *l2) {
        // write your code here
        ListNode result = ListNode(-1);
        ListNode *cur = &result;
        ListNode *digit;
        int sumNode;
        int incr = 0;
        while (l1 && l2) {
            sumNode = l1->val + l2->val + incr;
            if(sumNode < 10) {
                digit = new ListNode(sumNode);
                cur->next = digit;
                incr = 0;
            } else {
                digit =  new ListNode(sumNode - 10);
                cur->next  = digit;
                incr = 1;
            }
            l1 = l1->next;
            l2 = l2->next;
            cur = cur->next;
        }
        while (l1) {
           sumNode = l1->val + incr;
            if(sumNode < 10) {
                digit = new ListNode(sumNode);
                cur->next = digit;
                incr = 0;
            } else {
                digit =  new ListNode(sumNode - 10);
                cur->next  = digit;
                incr = 1;
            } 
            l1 = l1->next;
            cur = cur->next;
        }
        while (l2) {
           sumNode = l2->val + incr;
            if(sumNode < 10) {
                digit = new ListNode(sumNode);
                cur->next = digit;
                incr = 0;
            } else {
                digit =  new ListNode(sumNode - 10);
                cur->next  = digit;
                incr = 1;
            } 
            l2 = l2->next;
            cur = cur->next;
        }
        if(incr == 1) {
            digit = new ListNode (incr);
            cur->next = digit;
        }
        return result.next;
    }
};

A more clean code.

class Solution {
public:
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    ListNode *addLists(ListNode *l1, ListNode *l2) {
        // write your code here
        ListNode result = ListNode(-1);
        ListNode *cur = &result;
        ListNode *digit;
        int sumNode;
        int incr = 0;
        while (l1 || l2) {
            int num1 = l1 ? l1->val : 0;
            int num2 = l2 ? l2->val:0;
            sumNode = num1 + num2 + incr;
            digit = new ListNode(sumNode%10);
            cur->next = digit;
            cur = cur->next;
            incr = sumNode/10;
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;

        }

        if(incr == 1) {
            digit = new ListNode (incr);
            cur->next = digit;
        }
        return result.next;
    }
};