2. Add Two Numbers - Medium

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
Explanation: 342 + 465 = 807.

 

 

用一个dummy node记录相加后的结果，同时还需要一个node记录dummy linked list的头，用以最后return

在循环里，用sum和remainder分别表示两位的加和、进位值。注意最后如果进位值不为0，还需要增加一个新的listnode

注意dummy linked list的处理：dummy.next = new listnode(xxx), dummy = dummy.next

time: O(n), space: O(n)

/**
 * 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) {
        int remainder = 0;
        ListNode dummy = new ListNode(0);
        ListNode prehead = dummy;
        while(l1 != null || l2 != null) {
            int sum = remainder;
            if(l1 != null)  {
                sum += l1.val;
                l1 = l1.next;
            }
            if(l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            ListNode tmp = new ListNode(sum % 10);
            remainder = sum / 10;
            dummy.next = tmp;
            dummy = dummy.next;
        }
        if(remainder != 0) {
            dummy.next = new ListNode(remainder);
        }
        return prehead.next;
    }
}

 

二刷：

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode cur = dummy;
        int carry = 0;
        while(l1 != null || l2 != null) {
            if(l1 != null) {
                carry += l1.val;
                l1 = l1.next;
            }
            if(l2 != null) {
                carry += l2.val;
                l2 = l2.next;
            }
            cur.next = new ListNode(carry % 10);
            cur = cur.next;
            carry /= 10;
        }
        if(carry != 0) {
            cur.next = new ListNode(carry);
        }
        return dummy.next;
    }
}