2.两数相加
luffy 4/11/2023 力扣
# 1 两数之和
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。
示例 1:
输入:l1 = [2,4,3], l2 = [5,6,4]
输出:[7,0,8]
解释:342 + 465 = 807.
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示例 2:
输入:l1 = [0], l2 = [0]
输出:[0]
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示例 3:
输入:l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出:[8,9,9,9,0,0,0,1]
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解题思路:创建两个节点head、temp,遍历链表判断是否为空,将数值相加后num%10存在新的链表中,再将数值num/10,大于十的部分保留下次循环加上
java
/**
* 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 = new ListNode(-1),temp = head;
int num = 0;
while(l1 != null || l2 != null || num!=0){
if(l1 != null){
num+=l1.val;
l1 = l1.next;
}
if(l2 !=null){
num+=l2.val;
l2 = l2.next;
}
temp.next = new ListNode(num%10);
temp = temp.next;
num/=10;
}
return head.next;
}
}
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js
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
let head = new ListNode(-1);
let temp = head;
let num = 0;
while(l1 != null||l2 !=null || num !=0){
if(l1 != null){
num+=l1.val;
l1 = l1.next;
}
if(l2 != null){
num += l2.val;
l2 = l2.next;
}
temp.next = new ListNode(num%10);
temp = temp.next;
num = parseInt(num/10);
}
return head.next;
};
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Ts
/**
* Definition for singly-linked list.
* class ListNode {
* val: number
* next: ListNode | null
* constructor(val?: number, next?: ListNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
* }
*/
function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
let head:ListNode = new ListNode(-1);
let temp:ListNode = head;
let num:number = 0;
while(l1 != null || l2 !=null || num != 0){
if(l1 != null){
num += l1.val;
l1 = l1.next;
}
if(l2 != null){
num+=l2.val;
l2 = l2.next;
}
temp.next = new ListNode(num%10);
temp = temp.next;
num = num/10 | 0
// num = Math.floor(num/10);
}
return head.next;
};
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