Add Two Numbers

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
Explanation: 342 + 465 = 807.

Tags: Linked List, Math

思路

题意我也是看了好久才看懂，就是以链表表示一个数，低位在前，高位在后，所以题中的例子就是 342 + 465 = 807，所以我们模拟计算即可。

/**
 * 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) {
        ListNode node = new ListNode(0);
        ListNode n1 = l1, n2 = l2, t = node;
        int sum = 0;
        while (n1 != null || n2 != null) {
            sum /= 10;
            if (n1 != null) {
                sum += n1.val;
                n1 = n1.next;
            }
            if (n2 != null) {
                sum += n2.val;
                n2 = n2.next;
            }
            t.next = new ListNode(sum % 10);
            t = t.next;
        }
        if (sum / 10 != 0) t.next = new ListNode(1);
        return node.next;
    }
}

class Solution {
    fun addTwoNumbers(l1: ListNode?, l2: ListNode?): ListNode? {
        var n1 = l1
        var n2 = l2
        var firstNode: ListNode? = null
        var prev: ListNode? = null

        var co = 0
        while (n1 != null || n2 != null) {
            val van1 = n1?.`val` ?: 0
            val van2 = n2?.`val` ?: 0

            var item = van1 + van2 + co
            if (item >= 10) {
                item %= 10
                co = 1
            } else {
                co = 0
            }
            if (firstNode == null) {
                firstNode = ListNode(item)
                prev = firstNode
            } else {
                val newNode = ListNode(item)
                prev!!.next = newNode
                prev = newNode
            }

            n1 = n1?.next
            n2 = n2?.next
        }
        if (co == 1) {
            prev!!.next = ListNode(1)
        }
        return firstNode
    }
}

结语

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