LeetCode & linked list bug

add-two-numbers

shit test

/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
const log = console.log;
function ListNode(val, next) {
this.val = 0 || val;
this.next = null || next;
}
log(`l1`, l1)
log(`l2`, l2)
const arr1 = [];
const arr2 = [];
while(l1) {
arr1.push(l1.val);
l1 = l1.next;
}
while(l2) {
arr2.push(l2.val)
l2 = l2.next;
}
const sum = parseInt(arr1.reverse().join(``)) + parseInt(arr2.reverse().join(``));
// 807
const arr = Array.from(sum + ``).reverse().map(i => parseInt(i));
// [7, 0, 8]
const LinkedList = (value) => {
const node = new ListNode(value, ``);
if(!head) {
head = node;
} else {
let current = head;
while(current.next) {
current = current.next;
}
current.next = node;
}
};
let head = null;
for (let i = 0; i < arr.length; i++) {
LinkedList(arr[i]);
}
log(`head`, head)
return head;
// return arr;
}; /* 好垃圾的测试呀,为什么参数给的不是 ListNode, 而是 array! 答案没有毛病呀 l1 [2,4,3]
l2 [5,6,4]
head ListNode {
val: 7,
next: ListNode { val: 0, next: ListNode { val: 8, next: '' } }
} */

refs

https://leetcode.com/problems/add-two-numbers/

https://leetcode-cn.com/problems/add-two-numbers/submissions/

https://leetcode-cn.com/submissions/detail/99618022/



xgqfrms 2012-2020

www.cnblogs.com 发布文章使用:只允许注册用户才可以访问!


LeetCode & linked list bug的更多相关文章

  1. LeetCode Linked List Cycle II 和I 通用算法和优化算法

    Linked List Cycle II Given a linked list, return the node where the cycle begins. If there is no cyc ...

  2. LeetCode 全解(bug free 训练)

    1.Two Sum Given an array of integers, return indices of the two numbers such that they add up to a s ...

  3. [LeetCode] Linked List Random Node 链表随机节点

    Given a singly linked list, return a random node's value from the linked list. Each node must have t ...

  4. [LeetCode] Linked List Cycle II 单链表中的环之二

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Foll ...

  5. [LeetCode] Linked List Cycle 单链表中的环

    Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using ex ...

  6. [LeetCode]Linked List Cycle II解法学习

    问题描述如下: Given a linked list, return the node where the cycle begins. If there is no cycle, return nu ...

  7. LeetCode——Linked List Cycle

    Given a linked list, determine if it has a cycle in it. Follow up: Can you solve it without using ex ...

  8. LeetCode——Linked List Cycle II

    Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Foll ...

  9. [LeetCode] Linked List Components 链表组件

    We are given head, the head node of a linked list containing unique integer values. We are also give ...

随机推荐

  1. MySQL之谓词下推

    MySQL之谓词下推 什么是谓词 在SQL中,谓词就是返回boolean值即true或者false的函数,或是隐式转换为boolean的函数.SQL中的谓词主要有 LKIE.BETWEEN.IS NU ...

  2. oracle 常用语法()

    一ORACLE的启动和关闭 1在单机环境下 2在双机环境下 Oracle数据库有哪几种启动方式 1startup nomount 2startup mount dbname 3startup open ...

  3. CSS(简介or选择器)

    我们为什么需要CSS? 使用css的目的就是让网页具有美观一致的页面,另外一个最重要的原因是内容与格式分离 在没有CSS之前,我们想要修改HTML元素的样式需要为每个HTML元素单独定义样式属性,当H ...

  4. 解决java.lang.NoClassDefFoundError: ch/qos/logback/core/joran/spi/Pattern

    明明引入了这个,却提示没有 看下面文章: http://www.maocaoying.com/article/109

  5. 关于base64编码Encode和Decode编码的几种方式--Java

    Base64是一种能将任意Binary资料用64种字元组合成字串的方法,而这个Binary资料和字串资料彼此之间是可以互相转换的,十分方便.在实际应用上,Base64除了能将Binary资料可视化之外 ...

  6. Flink-v1.12官方网站翻译-P017-Execution Mode (Batch/Streaming)

    执行模式(批处理/流处理) DataStream API 支持不同的运行时执行模式,您可以根据用例的要求和作业的特点从中选择.DataStream API 有一种 "经典 "的执行 ...

  7. Codeforces Round #608 (Div. 2) E. Common Number (二分,构造)

    题意:对于一个数\(x\),有函数\(f(x)\),如果它是偶数,则\(x/=2\),否则\(x-=1\),不断重复这个过程,直到\(x-1\),我们记\(x\)到\(1\)的这个过程为\(path( ...

  8. hdu1541 Stars

    Problem Description Astronomers often examine star maps where stars are represented by points on a p ...

  9. 牛客编程巅峰赛S2第3场 Tree VI (树,dfs)

    题意:给你一个\(n\)个点的完全\(k\)叉树的先序遍历序列\(a\),还原这颗树并且求所有两个端点的异或和. 题解:用dfs在还原树的时候,把子节点和父亲节点的异或贡献给答案,对于每个节点,我们找 ...

  10. Codeforces Round #653 (Div. 3) A. Required Remainder (数学)

    题意:有三个正整数\(x,y,n\),再\(1\)~\(n\)中找一个最大的数\(k\),使得\(k\ mod\ x=y\). 题解:先记\(tmp=n/x\),再判断\(tmp*x+y\)的值是否大 ...