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LeetCode-Add Two Numbers

题目


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.

代码


/**
* Definition for singly-linked list.
* public class ListNode {
* public int val;
* public ListNode next;
* public ListNode(int val=0, ListNode next=null) {
* this.val = val;
* this.next = next;
* }
* }
*/
// solution 1
public class Solution {
public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = null;
ListNode tail = null;
var carry = 0;
while(l1 != null || l2 != null || carry != 0) {

var val1 = l1 == null ? 0 : l1.val;
var val2 = l2 == null ? 0 : l2.val;
var sum = val1 + val2 + carry;
carry = sum/10;
sum %= 10;

if (head == null) {
head = tail = new ListNode(sum);
}else {
tail.next = new ListNode(sum);
tail = tail.next;
}

if (l1 != null) {
l1 = l1.next;
}

if (l2 != null) {
l2 = l2.next;
}
}

return head;
}
}

// solution 2
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
var head = new ListNode();
var p = head;
var curry = 0;
while (l1 != null || l2 != null)
{
var val1 = l1 != null ? l1.val : 0;
var val2 = l2 != null ? l2.val : 0;
var sum = val2 + val1 + curry;
p.next = new ListNode(sum % 10);
p = p.next;
curry = sum / 10;
if (l1 != null)
{
l1 = l1.next;
}
if (l2 != null)
{
l2 = l2.next;
}
}
if (curry != 0)
{
p.next = new ListNode(curry);
}
return head.next;
}

// solution 3
public ListNode AddTwoNumbers(ListNode l1, ListNode l2)
{
var head = new ListNode();
var sum = l1.val + l2.val;
head.val = sum % 10;
head.val = sum % 10;
if (l1.next != null || l2.next != null || sum/10 != 0)
{
head.next = AddTwoNumbers(
l1.next == null ? new ListNode(sum / 10) : new ListNode(l1.next.val + sum / 10, l1.next.next),
l2.next ?? new ListNode());
}
return head;
}