import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * Source : https://oj.leetcode.com/problems/plus-one/ * * * Given a non-negative number represented as an array of digits, plus one to the number. * * The digits are stor…

题意:给一个数按位存放在一个int数组中,要求返回这个数加一后的数组. 懒人解法: public class Solution { public int[] plusOne(int[] digits) { java.math.BigInteger Bigdigits = new java.math.BigInteger(toString(digits)); String s = Bigdigits.add(new java.math.BigInteger("1")).toString(…

前言   [LeetCode 题解]系列传送门:  http://www.cnblogs.com/double-win/category/573499.html   1.题目描述 Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head…

问题描述:一个数组每一位代表一个数字的每一位.数字高位在数组的低位.求数字加1后得到新数组. 算法分析:要从数组的高位到低位进行遍历. public class PlusOne { public int[] plusOne(int[] digits) { int len = digits.length; int carry = 0; digits[len-1] += 1; if(digits[len-1] >= 10) { digits[len-1] = digits[len-1]-10; ca…

Given a non-negative number represented as a singly linked list of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. Example: Input: 1->2->3 Output: 1->2->4 这道题给了我们一个链表,用来模拟一…

Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 将一个数字的每个位上的数字分别存到一个一维向量中,最高位在最开头,我们需要给这个数字加一,即在末尾数字加一,如果末尾数字是9,那么则会有进位问题,而如果…

Array 448.找出数组中所有消失的数 要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间 思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…

利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectangle-in-histo…

(记忆线:当时一刷完是1-205. 二刷88道.下次更新记得标记不能bug-free的原因.)   88-------------Perfect Squares(完美平方数.给一个整数,求出用平方数来相加得到最小的个数) public class Solution{ public static void main(String[] args){ System.out.println(numSquares(8)); } public static int numSquares(int n){ //…

Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 题意:给定一个由数组表示的数字,加上一,返回由数组表示的结果 思路: 参考Discuss的解法,对于每一位来说,只有等于9的时候才会进位,小于就的时候,…

Google.Youtube.Facebook等均支持IPv6访问,IPv4网络的用户大部分都无法访问,比如Gmail,Google Docs等等各种相关服务.而该类网站大部分均已接入IPv6网络,因此通过IPv6访问则不受任何限制,尤其是对教育网用户来说,使用IPv6更是免费的服务. 而接入IPv6网络,也不仅仅只有教育网用户才能享受,普通公网用户也都可以通过隧道或软件来接入IPv6网络,但根据所处的网络环境.接入方式,访问IPv6的速度通常取决于本地网络宽带提供商与所接入的IPv6隧道服务器…

414. Third Maximum Number 给一个非空的整数数组,找到这个数组中第三大的值,如果不存在,那么返回最大的值.要求时间复杂度为o(n) 例如: Example 1: Input: [3, 2, 1] Output: 1 Explanation: The third maximum is 1. Example 2: Input: [1, 2] Output: 2 Explanation: The third maximum does not exist, so the maxi…

原题链接在这里:https://leetcode.com/problems/plus-one-linked-list/ 题目: Given a non-negative number represented as a singly linked list of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list.…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

现在才来了解java8,是不是后知后觉了点? 新的编程技术,个人不喜欢第一时间跟进. 待社区已有实践积淀再切入似乎更划算些? 一点点精明的考虑. 不多说,上代码. //读<写给大忙人看的java se 8>做的笔记代码 //希望对忙到连这书都没工夫看的你,匆匆一瞥,留下印象 //祝编程愉快 public class MainTest { //第一章,讲lambda表达式 //lambda表达式类似javascript的函数字面量,可用于替代java的匿名内部类 //基本型为 (形参列表)-&g…

Given a non-negative number represented as a singly linked list of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. Example: Input: 1->2->3 Output: 1->2->4 ListNode* plusOne…

# Copyright (c) 2014-2016, racaljk.# https://github.com/racaljk/hosts# Last updated: 2016-07-03 # This work is licensed under a CC BY-NC-SA 4.0 International License.# https://creativecommons.org/licenses/by-nc-sa/4.0/ # http://laod.cn/hosts/2016-goo…

刚才又人问我,要是群文件被鉴定为病毒那怎么下载? 我简单说一下吧: 其实qq客户端过滤比较严的,而web段却还是老一套,很多人说出现这个情况其实是腾讯已经把他库里面的文件删了,其实不然 如果源删了,那么web段也就不可能下载到了,咱们从这方面绕过它的验证(QQ群空间 http://qun.qzone.qq.com/group) 看图: 收工 扩展: 上传绕过:其实上传绕过太简单了,比如设置一个压缩包密码,比如压缩包里面套压缩包等等的,都是可以绕过的 赠送: 外网突破-打开就能用 http://p…

C#控制语句 控制语句: goto语句 If语句 do while循环  for循环  while循环 switch语句 三元运算符   <test?><resultIfTrue>:<resultIfFalse> foreach语句: 例如:int[] arr={1,2,3}; foreach(int I int arr) {Console.WriteLine(i);} 函数 两个关键字 static: 作用于变量:编译时分配内存,程序结束时释放 作用于函数:该函数只在…

Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 分析:思路是很清晰的,就是从数字最低位开始向上循环,如果是9就变成0,如果不是就直接当前位加1,返回.如果全都是9,就在最高位加一个1. 但是我写C++…

问题描述: Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 很简单就是像写大数加法那样,代码如下 class Solution { public: vector<int> plusOne(vecto…

Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. Have you met this question in a real interview? Yes Example Given [1,2,3] wh…

package plusOne66; /* Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. */ public class Solution { public static int[] plusOne…

hosts代理文件:C:\Windows\System32\drivers\etc\HOSTS 内容如下: # Copyright (c) -, racaljk. # https://github.com/racaljk/hosts # Last updated: -- # This work is licensed under a CC BY-NC-SA 4.0 International License. # https://creativecommons.org/licenses/by-n…

(1)Plus One 解题思路:模拟现实中做加法的方式,在个位加一,并考虑进位的情况.代码如下: public class Solution { public int[] plusOne(int[] digits) { int carries = 1; for (int i = digits.length - 1; i >= 0 && carries > 0; i--) { int sum = digits[i] + carries; digits[i] = sum % 10…

Find Kth Largest Number public class Solution { public int findKthLargest(int[] nums, int k) { return findK(nums,nums.length - k,0,nums.length-1); } private int findK(int[] nums,int k, int start, int end){ int parti = nums[start],i=start,m=start; for…

本文总结一下浏览器在 javascript 的加载方式. 关键词:异步加载(async loading),延迟加载(lazy loading),延迟执行(lazy execution),async 属性, defer 属性   一.同步加载与异步加载的形式   1. 同步加载 我们平时最常使用的就是这种同步加载形式: <script src="http://yourdomain.com/script.js"></script> 同步模式,又称阻塞模式,会阻止浏览…

题意让大数加1 我的做法是先让个位+1,再倒置digits,然后进位,最后倒置digits,得到答案. class Solution { public: vector<int> plusOne(vector<int> &digits) { digits[digits.size() -]++; //个位+1 reverse(digits.begin(),digits.end());//倒置digits ; i < digits.size() - ; ++i){//除了最…

leetcode代码 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectang…

Plus One https://leetcode.com/problems/plus-one/ Given a non-negative number represented as an array of digits, plus one to the number. The digits are stored such that the most significant digit is at the head of the list. 算法描述 这个题目需要注意的地方是假设数组的长度是n,…