题目:一个int的array,除了一个元素只有一个其余的都是两个,找到这一个的元素. 使用:逻辑运算 XOR异或运算 关于逻辑运算的总结[转] &&和||:逻辑运算符 &和|:按位运算符 &&是且的意思,a&&b 两者都为真才为真. ||是或的意思,a||b 两者有一为真即真. &,|是位运算符.即对位进行运算, 如00000011 & 00000001=00000001 00000011 | 00000001=00000011 对于…

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? 题意:给定数组,除一个数仅出现一次外,其余都出现三次,找到仅出现一次的数字…

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)…

[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…

问题: 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?   Single Number I 升级版,一个数组中其它数出现了…

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…

Single Number I : 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? Solution: 解法不少,贴一种: class…

Given an array of integers, every element appears twice except for one. Find that single one. 思路: 最经典的方法,利用两个相同的数异或结果为0的性质,则将整个数组进行异或,相同的数俩俩异或,最后得到的就是那个single number,复杂度是O(n) 代码: class Solution { public: int singleNumber(vector<int>& nums) { int…