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? 题意:给定一未排序的数组,其中只有一个数仅出现一次,其余都出现两次,找到仅出现一次的这…

题意:给一个序列,序列中只有1个是单个的,其他都是成对出现的.也就是序列中有奇数个元素.要求找出这个元素. 思路:成对出现用异或最好了.两个同样的数一异或就变零,剩下的,就是那个落单的. class Solution { public: int singleNumber(vector<int>& nums) { ; ; i<nums.size(); i++) res^=nums[i]; return res; } }; AC代码…

题意:给一个序列,其中只有1个元素只出现1次,其他的都一定出现3次.问这个出现一次的元素是多少? 思路: (1)全部元素拆成二进制,那么每个位上的1的个数应该是3的倍数,如果不是3的倍数,则ans的这个位肯定为1. 22ms class Solution { public: int singleNumber(vector<int>& nums) { ]={}; ; i<nums.size(); i++) ; j<=; j++) &(nums[i]>>j)…

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…

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…

[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…

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)…

题目地址:here 题目大意:一个整数数组中,只有一个数出现一次,其余数都出现3次,在O(n)时间,O(1)空间内找到这个出现一次的数 对于”只有一个数出现一次,其余数出现2次“的情况,很简单,只要把数组中所有数异或的结果就是这个出现一次的数,leetcode上也有这个题目. 关于本题:我们用一个大小为32的数组cnt来记录:(int数据的每个二进制位1的出现次数)%3,程序中cnt[i]记录第i+1个二进制位.如果某个数出现了三次并且他的第i个二进制位为1,因为做了模3操作,那么该数对数组cn…