Given 2*n + 1 numbers, every numbers occurs twice except one, find it. Have you met this question in a real interview? Yes Example Given [1,2,2,1,3,4,3], return 4 Challenge One-pass, constant extra space. LeetCode上的原题,请参见我之前的博客Single Number. class So…

C++ (1)异或操作 3^3=0 (2)for (auto &i : Obejuct) {} class Solution { public: /** * @param A: Array of integers. * return: The single number. */ int singleNumber(vector<int> &A) { // write your code here ; for (auto &i : A) { ret ^= i; } retu…

Given 2*n + 2 numbers, every numbers occurs twice except two, find them. Example Given [1,2,2,3,4,4,5,3] return 1 and 5 Challenge O(n) time, O(1) extra space. 利用bitwise XOR的特点,n个数(0或1),如果1的个数为奇数,则n个数bitwise XOR结果为1,否则为0 先将所有的数异或,得到的将是x和y以后之后的值n. 找到这个…

一篇解析比较详细的文章:http://www.acmerblog.com/leetcode-single-number-ii-5394.html C++ 解法(1) 求出每个比特位的数目,然后%3,如果这个比特位只出现1次,那么这比特位就会余1,否则就会整除. 把每个余下的比特位求出来,就知道是哪个数只出现1次了. class Solution { public: /** * @param A : An integer array * @return : An integer */ int si…

Single Number III Given 2*n + 2 numbers, every numbers occurs twice except two, find them. Example Given [1,2,2,3,4,4,5,3] return 1 and 5 Challenge O(n) time, O(1) extra space. 可耻的百度了..点这里 还是好理解的.当然不一定要用最低位的1,任意找一个1即可.不过有lowbit = xor & -xor 这个公式当然是这个…

Single Number II Given 3*n + 1 numbers, every numbers occurs triple times except one, find it. Example Given [1,1,2,3,3,3,2,2,4,1] return 4 Challenge One-pass, constant extra space. 统计每一位上的1出现的次数,然后模3 , 题目上的3 * n + 1给了提示,然后又做过一题2 * n + 1的位操作. public…

LintCode 83. Single Number II (Medium) LeetCode 137. Single Number II (Medium) 以下算法的复杂度都是: 时间复杂度: O(n) 空间复杂度: O(1) 解法1. 32个计数器 最简单的思路是用32个计数器, 满3复位0. class Solution { public: int singleNumberII(vector<int> &A) { int cnt[32] = {0}; int res = 0; f…

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)…