http://www.acmerblog.com/leetcode-single-number-ii-5394.html acm之家的讲解总是那么到位 public class Solution { public int singleNumber(int[] A) { int ans=0; //0^X=X x^X=x for(int i=0;i<A.length;i++) { ans=ans^A[i]; } return ans; } } sing number (2) 高端方法太难了,不想看了…
LeeCode是一个有意思的编程网站,主要考察程序员的算法 第二题: You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked l…
运行/sbin/shutdown now最后显示:Telling INIT to go to single user mode.INIT:Going single userINIT:Sending grocesses the TERM signalINIT:no more processes left in this runlevelsh-3.2#在此状态下运行csh后自动登录为root.若运行exit系统就重新启动回到登录画面. 登录后运行reboot,系统可正常重启. 解决方法: 1.…
Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5]. Note: The order…
Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 这道题是之前那道 Single Number 单独的数字的延伸,那道题的解法…
Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 本来是一道非常简单的题,但是由于加上了时间复杂度必须是O(n),并且空间复杂度为O(1)…
学习这件事在任何时间都不能停下.准备坚持刷leecode来提高自己,也会把自己的解答过程记录下来,希望能进步. Two Sum Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution. Example: Given n…
[1]LeetCode 136 Single Number 题意:奇数个数,其中除了一个数只出现一次外,其他数都是成对出现,比如1,2,2,3,3...,求出该单个数. 解法:容易想到异或的性质,两个相同的数异或为0,那么把这串数从头到尾异或起来,最后的数就是要求的那个数. 代码如下: class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.size();i++) sum ^= nums[i…