Add Two Numbers

Description

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.

Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.

Link

Lintcode_ladder

Method

1. x
2. x

Example

1. 1

/**
 * 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) {
    // ------------------------------------------------
    // Simplified version
        if (!l1) {
            return l2;
        } else if (!l2){
            return l1;
        }
        ListNode dummy(0);
        ListNode* prev = &dummy;
        int sum = 0;
        int carry = 0;
        while (l1 || l2) {
            // get sum
            // l1 && l2 are not empty
            if (l1 && l2) {
                sum = l1->val + l2->val + carry;
                l1 = l1->next;
                l2 = l2->next;
            } else if (l1) {
                // only l1 is not empty
                sum = l1->val + carry;
                l1 = l1->next;
            } else {
                // only l2 is not empoty
                sum = l2->val + carry;
                l2 = l2->next;
            }
            // update result
            carry = sum / 10;
            sum %= 10;
            prev->next = new ListNode(sum);
            prev = prev->next;
        }
        if (carry > 0) {
            prev->next = new ListNode(carry);
        }
        return dummy.next;
    // ----------------------------------------------------
    // naive version
        if (!l1) {
            return l2;
        } else if (!l2){
            return l1;
        }
        ListNode dummy(0);
        ListNode* prev = &dummy;
        int sum = 0;
        int carry = 0;
        while (l1 || l2) {
            if (l1 && l2) {
                sum = l1->val + l2->val + carry;
                carry = sum / 10;
                sum %= 10;
                prev->next = new ListNode(sum);
                l1 = l1->next;
                l2 = l2->next;
                prev = prev->next;
            } else if (l1) {
                sum = l1->val + carry;
                carry = sum / 10;
                sum %= 10;
                prev->next = new ListNode(sum);
                l1 = l1->next;
                prev = prev->next;
            } else if (l2) {
                sum = l2->val + carry;
                carry = sum / 10;
                sum %= 10;
                prev->next = new ListNode(sum);
                l2 = l2->next;
                prev = prev->next;
            }
        }
        if (carry > 0) {
            prev->next = new ListNode(carry);
        }
        return dummy.next;
    }
};

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