LetCode算法--2.两数相加

本文主要是介绍LetCode算法--2.两数相加，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

给你两个 非空 的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储 一位 数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/add-two-numbers

写法一：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode listNode = new ListNode();
        ListNode carry = listNode;
        int x;
        int y;
        int arr = 0;


        while (l1 != null || l2 != null) {
            if (l1 != null) {
                x = l1.val;
            } else {
                x = 0;
            }
            if (l2 != null) {
                y = l2.val;
            } else {
                y = 0;
            }

            int sum = x + y + arr;

            carry.next = new ListNode(sum % 10);
            carry = carry.next;
            arr = sum/10;

            if(l1 != null) l1 =l1.next;
            if (l2 != null) l2 = l2.next;

            if (arr != 0){
                carry.next = new ListNode(arr);
            }
        }
        return listNode.next;
    }
}

写法二：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    
   public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode listNode = new ListNode();
        ListNode carry = listNode;
        int arr = 0;
        
        while (l1 != null || l2 != null){
            int x = l1 == null? 0: l1.val;
            int y = l2 == null? 0: l2.val;

            int sum = x + y + arr;
            carry.next = new ListNode(sum%10);
            carry = carry.next;
            arr = sum/10;
            
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
            
            if (arr != 0) carry.next = new ListNode(arr);
        }
        return listNode.next;
    }
}

写法三：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    
  public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = null, tail = null;
        int carry = 0;
        while (l1 != null || l2 != null){
            int x = l1 == null ? 0:l1.val;
            int y = l2 == null ? 0:l2.val;
            int sum = x + y + carry;

            if (head == null){
                head = tail = new ListNode(sum%10);
            }else {
                tail.next = new ListNode(sum%10);
                tail = tail.next;
            }
            carry = sum / 10;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        if (carry != 0){
            tail.next = new ListNode(carry);
        }
        return head;
    }
}

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