LetCode算法--2.两数相加
2022/8/29 14:25:35
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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